ENM 503 Block 2 Lesson 7 – Matrix Methods Everything you would want to know about The Matrix and then some… this way Narrator: Charles Ebeling
ENM 503 Block 2 Lesson 7 – Matrix Methods Everything you would want to know about The Matrix and then some… this way Narrator: Charles Ebeling.
Dec 27, 2015
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ENM 503 Block 2Lesson 7 ndash Matrix Methods
Everything you would want to know about The Matrix and
then somehellip
this way
Narrator Charles Ebeling
Applications Solving systems of linear equations Regression analysis Markov processes Linear programming Nonlinear optimization Queuing Reliability Inventory ndash MRP systems
These only begin to show the potential
of the matrix
Matrix and Vectors
A matrix is a rectangular array of elements which are operated on as a single object The elements are often numbers but could be any mathematical object provided that it can be added and multiplied with acceptable properties
Vectors are strongly related to matrices they can be considered as a matrix having only a single row (row vector) or a single column (column vector)
Examples
1
2
3
11 12 13
21 22 23
31 32 33
34 65
12 23
8 0
y
a b c d y
y
a a a
a a a
a a a
X Y
A B
X is a 1 x 4 row vector Y is a 3 x 1 column vector
A is a 3 x 3 matrix and B is a 3 x 2 matrix
An m x n Matrix
11 1
1
n
ij
m mn
a a
a
a a
A
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Applications Solving systems of linear equations Regression analysis Markov processes Linear programming Nonlinear optimization Queuing Reliability Inventory ndash MRP systems
These only begin to show the potential
of the matrix
Matrix and Vectors
A matrix is a rectangular array of elements which are operated on as a single object The elements are often numbers but could be any mathematical object provided that it can be added and multiplied with acceptable properties
Vectors are strongly related to matrices they can be considered as a matrix having only a single row (row vector) or a single column (column vector)
Examples
1
2
3
11 12 13
21 22 23
31 32 33
34 65
12 23
8 0
y
a b c d y
y
a a a
a a a
a a a
X Y
A B
X is a 1 x 4 row vector Y is a 3 x 1 column vector
A is a 3 x 3 matrix and B is a 3 x 2 matrix
An m x n Matrix
11 1
1
n
ij
m mn
a a
a
a a
A
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Matrix and Vectors
A matrix is a rectangular array of elements which are operated on as a single object The elements are often numbers but could be any mathematical object provided that it can be added and multiplied with acceptable properties
Vectors are strongly related to matrices they can be considered as a matrix having only a single row (row vector) or a single column (column vector)
Examples
1
2
3
11 12 13
21 22 23
31 32 33
34 65
12 23
8 0
y
a b c d y
y
a a a
a a a
a a a
X Y
A B
X is a 1 x 4 row vector Y is a 3 x 1 column vector
A is a 3 x 3 matrix and B is a 3 x 2 matrix
An m x n Matrix
11 1
1
n
ij
m mn
a a
a
a a
A
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Examples
1
2
3
11 12 13
21 22 23
31 32 33
34 65
12 23
8 0
y
a b c d y
y
a a a
a a a
a a a
X Y
A B
X is a 1 x 4 row vector Y is a 3 x 1 column vector
A is a 3 x 3 matrix and B is a 3 x 2 matrix
An m x n Matrix
11 1
1
n
ij
m mn
a a
a
a a
A
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
An m x n Matrix
11 1
1
n
ij
m mn
a a
a
a a
A
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Vector Matrix Operations Vectors and matrices can be added (or subtracted) and
multiplied when their dimensions are in agreement
To add or subtract two vectors or two matrices having the same dimensions just add their corresponding elements
A plusmn B = aij plusmn bij
To multiply two vectors multiply corresponding elements and add The result is a scalar (dot product) Both vectors must have the same number of elements
1
n
i ii
x y
XY
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Vector Example
3 8 2 7 3 6 1 2
3 8 2 7 3 6 1 2 6 14 3 9
3 3 8 6 2 1 7 2 73x x x x
X Y
X Y
X Y
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Example Matrix Addition and scalar multiplication
5 6 0 6 7 1
2 1 5 0 2 3
9 0 3 0 3 2
11 13 1
2 3 8
0 3 1
5 6 0
2 5
9 0 3
A B
A B B A
A scalar multiplication
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Matrix Multiplication If A is an m x n matrix and B is an n x p matrix
then C = A x B is an m x p matrix where
The ij element of C is found by multiplying the ith row of A times the jth column of B (equivalent to a vector multiplication)
1
n
ij ik kjk
c a b
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Example Matrix Multiplication
3 52 1 4
2 13 0 2
4 2
2 3 3 2
6 2 16 10 1 8 24 17
9 0 8 15 0 4 1 11
x x
x
A B
A B
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Properties of Matrix Operations
A (BC) = (AB) C
A (B+C) = AB + AC
(B+C) A = BA + CA
however A B B A (both are defined only if A and Bare n x n matrices)
and A A = A2 (only if a square matrix ie dimension n x n)
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
An interesting sidelight
A B = 0 does not necessarily imply that A = 0 or B = 0
For example1 1 1 1 0 0
2 2 1 1 0 0
Yes that is really
interesting
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Transpose
If A = ajk then At = akj
Each row of A becomes a column of At
2 32 1 4
1 03 0 2
4 2
2 4 7 2 4 7
4 8 1 4 8 1
7 1 5 7 1 5
t
t
A A
A A
If A = At then Ais a symmetric matrixie aij = aji
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Properties of the Transpose
(At)t = A
(A + B)t = At + Bt
(kA)t = k At
(AB)t = Bt At
Quick student exercise Show that AtA is symmetric using the above properties
Quick student exercise Create an example to illustrate each property
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Some Special Matrices1 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0
1 0
0 0 0 0 0 1 0 0 0 0 0 0
I Ο
The Identity Matrix (n x n) The Null Matrix
AI = IA = A OA = AO = O
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
More Special Matrices
11 12 1 11
22 21 22
33 33
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
n
nn mn nn
a a a a
a a a
a a
a a a
A B
Upper triangular Lower Triangular
all zeros
Now ainrsquot that special
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Diagonal Matrix
11
22
33
0 0 0 0 0
0 0 0 0 0
0
0
0 0 0 0 0 nn
a
a
a
a
A
main diagonal
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Determinant
For a square (n x n) matrix A the determinant is defined as a scalar computed from the sum of n terms of the form ( a1i a2j hellip anr) the sign alternating and depending upon the permutation
11
1 2all nterms
1
( )
in
i j nr
n nn
a a
a a a
a a
A
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
A 2 x 2 Determinant
det( ) | |
5 45 3 4 2 7
2 3
a bad bc
c d
A A
5 4
2 3
a b
c d
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
A 3 x 3 determinant
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 13 32 21
31 22 13 21 12 33 11 23 32
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
+
-
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Properties of Determinants1 |A| = |At|
2 If ajk = 0 for all k or for all j |A| = 0
3 Interchange any 2 rows Arsquo |A| = - |Arsquo|
4 For scalar k |kA| = k |A|
5 If there are 2 identical rows or columns |A| = 0
6 |AB| = |A| |B|
7 If A is triangular 1
n
jjj
a
A
Quick student exercise Create an example to illustrate each property
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Cofactors Minor ndash determinant of order (n-1) obtained
by removing the jth row and kth column of A
Cofactor (-1)j+k Minorjk = Ajk
Cofactor matrix - A matrix with elements that are the cofactors term-by-term of a given square matrix ndash [Ajk]
Adjoint Matrix = transpose of the cofactor matrix ndash [Ajk]t
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Example Cofactor Matrix
11 12 13
21 22 23
31 32 33
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
4 5 0 5 0 424 5 4
0 6 1 6 1 0
2 3 1 3 1 212 3 2
0 6 1 6 1 0
2 3 1 3 1 22 5 4
4 5 0 5 0 4
A A A
A A A
A A A
A
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Adjoint Matrix
1 2 3 24 5 4
0 4 5 cofactor matrix 12 3 2
1 0 6 2 5 4
24 12 2
Adjoint 5 3 5
4 2 4
t
jk
jk
A A
A
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Expansion by Cofactors (2 x 2)
1
n
jk jkk
a A
A
11 12 2 311 22 12 21
21 22
11 22 21 12
( 1) | | ( 1) | |a a
a a a aa a
a a a a
I call this technique the Laplace expansion
Pierre-Simon LaplaceBorn 23 March 1749 in Normandy FranceDied 5 March 1827 in Paris France
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
11 12 13
21 22 23
31 32 33
2 3 422 23 21 23 21 2211 12 13
32 33 31 33 31 32
1 1 1
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
Quick student exercise Complete the example below(ie express algebraically)
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Example Expansion by Cofactors (3x3)
1
n
jk jkk
a A
A
1 2 33 1 2 1 2 3
2 3 1 1 2 31 2 3 2 3 1
3 1 2
1(5) 2(1) 3( 7) 18
Expanding about row 1
Quick student exercise Expand about column 2and show that the sameresult is obtained
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Matrix Inverse
A square matrix A may have an inverse matrix A-1 such that
If such a matrix exists then A is said to be nonsingular or invertible The inverse matrix A-1 will be unique
A square matrix A is said to be singular if |A| = 0If |A| 0 then A is said to be nonsingular
-1 -1AA A A I
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Necessary Example3 2 5 5
then1 2 25 75
3 2 5 5 1 0since
1 2 25 75 0 1
-1
-1
A A
AA
How did you ever find A-1 A lucky guess or somethin
Quick student exercise
Show A-1 A = I for thisexample
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
(we will come back to this problemand solve it shortly)
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Properties of Inverses
1 1 1
11
1 1 tt
AB B A
A A
A A
How much more of this
can I absorb
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding Inverses Method 1 ndash Adjoint Matrix
Method 2 ndash Gauss-Jordan Elimination Method Elementary Row Operations (ERO)
define an augment matrix [AI] where I is an n x n identity matrix
Perform ERO on [AI] to obtain [IA-1]
1
t
jk A
AA
Did you know If |A| = 0 then A-1 does not exist
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Method 1 The Adjoint Method
1
1 2 3 24 12 2
0 4 5 Adjoint 5 3 5 22
1 0 6 4 2 4
24 12 2
5 3 5109 5454 0909
4 2 42273 1363 2273
221818 0909 1818
A
A A
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Method 2 ndash The Gauss-Jordan Way
Carl Friedrich Gauss1777-1855
Wilhelm Jordan 1838-1922
This is our way of doing it
We do it with elementary row
operations
Gaussian Elimination
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Elementary Row Operations (ERO)
1) Interchange ith and jth row Ri Rj
2) Multiply the ith row by a nonzero scalar
Ri kRi
3) Replace the ith row by k times the jth row plus the ith row
Ri kRj + Ri
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Augmented Matrix
[ A I ]
[ I A-1]
EROrsquos
need an example
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Example2 2 3 1 0 0 1 1 3 2 1 2 0 0
1 3 1 0 1 0 1 3 1 0 1 0
4 1 2 0 0 1 4 1 2 0 0 1
1 1 3 2 1 2 0 0 1 1 3 2 1 2 0 0
0 2 1 2 1 2 1 0 0 2 1 2 1 2 1 0
4 1 2 0 0 1 0 3 4 2 0 1
1 1 3 2 1 2 0 0 1 0 7 4 3 4 1 2 0
0 1 1 4 1 4 1 2 0
0 3 4 2 0 1
0 1 1 4 1 4 1 2 0
0 0 19 4 11 4 3 2 1
1 0 7 4 3 4 1 2 0 1 0 0 519 119 7 19
0 1 1 4 1 4 1 2 0 0 1 0 2 19 819 119
0 0 1 1119 6 19 4 19 0 0 1 1119 6 19 4 19
2 2 3
1 3 1
4 1 2
A 1
5 1 71
2 8 119
11 6 4
A
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Matrices and Systems of Linear Equations
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m n n m
a a a x b
a a a x b
a a a x b
A x b
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
Ax = b
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
The Augmented Matrix
Ax = b
[ A I b ]
[ I A-1 brsquo ]
x = brsquo
EROrsquos
-1 -1
xΑΙ b
0
xI A A b b
0
Did you know implied in A-1 are all the EROrsquos (arithmetic) to go from A to A-1
need an example
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
An Example3 5 1 0 1
1 2 0 1 2
1 5 3 1 3 0 1 3
1 2 0 1 2
1 5 3 1 3 0 1 3
0 1 3 1 3 1 5 3
1 5 3 1 3 0 1 3
0 1 1 3 5
1 0 2 5 8
0 1 1 3 5
R1 R1 3
R2 R2 ndash R1
R2 3 R2
R1 R1 ndash (53)R2
3 5 1
2 2
x y
x y
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Solving systems of linear eqs using the matrix inverse
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
11
2
mn
xb
x
bx
X b
11 1
1
n
m mn
a a
a a
A
Matrix solutionAX = bA-1 (AX) = (A-1A)X = IX =X = A-1b
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Example System or Eqs
3 5 1 3 5 1
2 2 1 2 2
2 5
1 3
2 5 1 8
1 3 2 5
x y x
x y y
x
y
-1
-1
A
x A b
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Cramerrsquos Rule for solving systems of linear equations
Gabriel CramerBorn 31 July 1704 in Geneva SwitzerlandDied 4 Jan 1752 in Bagnols-sur-Cegraveze France
Given AX = bLet Ai = matrix formed by replacing the ith column with b then
1 2ix i n iA
A
I do it with determinants
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
An Example of Cramerrsquos Rule
2 3 7 2 319
3 5 1 3 5
7 3 2 738 19
1 5 3 1
38 192 1
19 19
x y
x y
x y
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Cramer solving a 3 x 3 system2 3 2 1 1
1 1 1 1
2 3 4 1 2 3
( 6) (1) (2) ( 1) ( 3) ( 4) 5
3 1 1 2 3 1 2 1 3
1 1 1 10 1 1 1 5 1 1 1 0
4 2 3 1 4 3 1 2 4
10 5 02 1 0
5 5 5
x y z
x y z
x y z
x y z
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding A-1 for a 2 x 2
1 0
0 1
1
0
0
1
a b x y
c d z w
ax bz
ay bw
cx dz
cy dw
solve for x y z and w in terms of a b c and d
Letrsquos use Cramerrsquos rule
A rare moment ofinspiration among agroup of ENM students
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
1
0
b
d dx
a b ad bc
c d
1
0
a
c cz
a b ad bc
c d
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding A-1 for a 2 x 2
1
0
0
1
ax bz
cx dz
ay bw
cy dw
0
1
b
d by
a b ad bc
c d
0
1
a
c aw
a b ad bc
c d
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Finding A-1 for a 2 x 2
1
a bA
c d
d b
x y c aA
z w ad bc
I get it To find the inverse you swap the
two diagonal elements change the sign of the
two off-diagonal elements and divide by
the determinant
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
A Numerical Example
1
2 4
1 3
3 4
3 41 2
1 26 4
A
d b
c aA
ad bc
1 2 4 3 4 1 0
1 3 1 2 0 1AA
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Properties of Triangular Matrices
Triangular matrices have the following properties (prefix ``triangular with either ``upper or ``lower uniformly) The inverse of a triangular matrix is a triangular
matrix The product of two triangular matrices is a
triangular matrix The determinant of a triangular matrix is the
product of the diagonal elements A matrix which is simultaneously upper and lower
triangular is diagonal The transpose of a upper triangular matrix is a
lower triangular matrix and vice versa
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Determinants by Triangularization
4 1 1
5 2 0 16 0 15 0 10 0 21
0 3 2
4 1 1 4 1 1 4 1 1
5 2 0 0 3 4 5 4 0 3 4 5 4
0 3 2 0 3 2 0 0 7
4 1 1
0 3 4 5 4 (4)(3 4)(7) 21
0 0 7
Rrsquo2 -54 R1 + R2 Rrsquo3 -4 R2 + R3
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
Solving Systems of Equations the Easy Way
There must be an easier way
Why not use Excel with
VBA
to Excel with VBAhellip
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