Linear Algebra. Session 3 - Texas A&M Universityroquesol/Math_304... · Linear Algebra. Session 3 Dr. Marco A Roque Sol 01/30/2018 Dr. Marco A Roque Sol Linear Algebra. Session 3.

Matrices and Matrix AlgebraDeterminants I

Linear Algebra. Session 3

Dr. Marco A Roque Sol

01/30/2018

Dr. Marco A Roque Sol Linear Algebra. Session 3


Matrices, matrix algebra

Inverting diagonal matrices

Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible if and onlyif all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n

If D is invertible then D−1 = diag(d−11 , d−12 , · · · .d−1n )

d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n







d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix

D = diag(d1, d2, · · · .dn) is invertible if and onlyif all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn)

is invertible if and onlyif all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible

if and onlyif all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible if and onlyif

all diagonal entries are nonzero; di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible if and onlyif all diagonal entries are

nonzero; di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n





Theorem 3.1A diagonal matrix D = diag(d1, d2, · · · .dn) is invertible if and onlyif all diagonal entries are nonzero;

di 6= 0 for 1 ≤ i < n


d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n







d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n






If D

is invertible then D−1 = diag(d−11 , d−12 , · · · .d−1n )

d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n






If D is invertible

then D−1 = diag(d−11 , d−12 , · · · .d−1n )

d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n







d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n







d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n







d1 0 · · · 00 d2 · · · 0...

.... . .

...0 0 · · · dn

−1

=

d−11 0 · · · 0

0 d−12 · · · 0...

.... . .

...0 0 · · · d−1n




proof

If all di 6= 0,then clearly,

diag(d1, d2, · · · .dn)diag(d−11 , d−12 , · · · .d−1n ) =

diag(1, 1, · · · .1) = In

diag(d−11 , d−12 , · · · , d−1n )diag(d1, d2, · · · .dn) =

diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof

If

all di 6= 0,then clearly,


diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof

If all di 6= 0,

then clearly,


diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof


diag(d1, d2, · · · .dn)

diag(d−11 , d−12 , · · · .d−1n ) =

diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In

diag(d−11 , d−12 , · · · , d−1n )

diag(d1, d2, · · · .dn) =

diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now

suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that

di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0,

for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i .

Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then

for any n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any

n × n matrixB, the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB,

the ith row of the matrix DB is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB

is zero. Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero.

Hence DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In

Now suppose that di = 0, for some i . Then for any n × n matrixB, the ith row of the matrix DB is zero. Hence

DB 6= In. �




proof



diag(1, 1, · · · .1) = In


diag(1, 1, · · · .1) = In





Inverting 2× 2 matrices

The determinant of a 2× 2 matrix

A =

(a bc d

)is denoted by det(A) and defined by det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d

)is invertible if and only if det(A) 6= 0 . If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1





The determinant

of a 2× 2 matrix

A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d

)

is denoted by det(A) and defined by det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d

)is denoted by

det(A) and defined by det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d

)is denoted by det(A) and

defined by det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d

)is denoted by det(A) and defined by

det(A) = ad − bc

Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d

)

is invertible if and only if det(A) 6= 0 . If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d

)is invertible

if and only if det(A) 6= 0 . If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d

)is invertible if and only if

det(A) 6= 0 . If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d

)is invertible if and only if det(A) 6= 0 .

If det(A) 6= 0, then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d

)is invertible if and only if det(A) 6= 0 . If det(A) 6= 0,

then

A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1






A =

(a bc d


Theorem 3.2

A matrix

A =

(a bc d


A−1 =

(a bc d

)−1=

1

ad − bc

(d −b−c a

)−1Dr. Marco A Roque Sol Linear Algebra. Session 3



proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc

)In the case det(A) = ad − bc 6= 0 we have A−1 = [det(A)]−1B

In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B

the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)

then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc

)

In the case det(A) = ad − bc 6= 0 we have A−1 = [det(A)]−1B





proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc

)In the case

det(A) = ad − bc 6= 0 we have A−1 = [det(A)]−1B





proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc

)In the case det(A) = ad − bc 6= 0

we have A−1 = [det(A)]−1B





proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case

det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0

the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0

⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒

(A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0

⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒

(A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒

(I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒

B = 0⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0

⇒ B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒

B = 0, but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0,

but the zero matrix is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix

is singular !!!!! �




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc


In the case det(A) = ad − bc = 0 the matrix A is non invertible asotherwise⇒ AB = 0 ⇒ (A−1(AB) = A−10 = 0⇒ (A−1A)B = 0 ⇒ (I2B = 0 ⇒ B = 0⇒ B = 0, but the zero matrix is singular !!!!!

�




proof

Let B the matrix

B =

(d −b−c a

)then

AB = BA =

(ad − bc 0

0 ad − bc






System of n linear equations in n variables:a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2

...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn




System of n linear equations in n variables:

a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2

...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔

Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn





...am1x1 + am2x2 + · · ·+ amnxn = bm

⇔ Ax = b

where

A =

a11 a12 · · · a1na21 a22 · · · a2n

...an1 an2 · · · ann

x =

x1x2...xn

b =

b1b2...bn




Theorem 3.3

If the matrix A above, is invertible then the system has a uniquesolution, which is x = A−1b

General results on inverse matrices

Theorem 3.4

Given an n × n matrix A, the following conditions are equivalent:

(i) A is invertible.

(ii) x = 0 is the only solution of the matrix equation Ax = 0.

(iii) The matrix equation Ax = b has a unique solution for anyn-dimensional column vector b.




Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








Theorem 3.3



Theorem 3.4








(iv) The row echelon form of A has no zero rows.has a unique solution for any n-dimensional column vector b

(v) The reduced row echelon form ofA is the identity matrix.

Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5







Theorem 3.5







Theorem 3.5







Theorem 3.5

Suppose

that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of

elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations

converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A

into the identity matrix. Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix.

Then the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then

the same sequence ofoperations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence of

operations converts the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts

the identity matrix into the inverse matrix A−1






Theorem 3.5

Suppose that a sequence of elementary row operations converts amatrix A into the identity matrix. Then the same sequence ofoperations converts the identity matrix

into the inverse matrix A−1






Theorem 3.5







Theorem 3.5





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always has a unique solution. Also, in the formercase the reduced row echelon form of A is I .




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a unique solution. Also, in the formercase the reduced row echelon form of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution.

Also, in the formercase the reduced row echelon form of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also,

in the formercase the reduced row echelon form of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also, in the formercase

the reduced row echelon form of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also, in the formercase the reduced

row echelon form of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also, in the formercase the reduced row echelon form

of A is I .




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Hence the row echelon form of a square matrix A is either stricttriangular or else it has a zero row. In the former case, theequation Ax = b always has a unique solution. Also, in the formercase the reduced row echelon form of A is

I .




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Example 3.2

Check whether the matrix A,given by 3 −2 11 0 1−2 3 0

is invertible.

Solution

We convert it to row echelon form.

Interchange the 1st row with the 2nd row: 1 0 13 −2 1−2 3 0

Add −3 times the 1st row to the 2nd row: 1 0 1

0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2

Check

whether the matrix A,given by 3 −2 11 0 1−2 3 0

is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2

Check whether the matrix A,

given by 3 −2 11 0 1−2 3 0

is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2

Check whether the matrix A,given by

3 −2 11 0 1−2 3 0

is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution

We convert it

to row echelon form.



0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution


Interchange the 1st row with the 2nd row:

1 0 13 −2 1−2 3 0


0 −2 −3−2 3 0




Example 3.2


is invertible.

Solution



Add −3 times the 1st row to the 2nd row:

1 0 10 −2 −3−2 3 0




Example 3.2


is invertible.

Solution




0 −2 −3−2 3 0




Add 2 times the 1st row to the 3rd row : 1 0 10 −2 −30 3 2

Multiply the 2nd row by −1/2 : 1 0 1

0 1 3/20 3 2




Add 2 times the 1st row to the 3rd row :

1 0 10 −2 −30 3 2


0 1 3/20 3 2





Multiply the 2nd row by −1/2 : 1 0 10 1 3/20 3 2





Multiply the 2nd row by −1/2 :

1 0 10 1 3/20 3 2






0 1 3/20 3 2




Add −3 times the 2nd row to the 3rd row: 1 0 10 1 3/20 0 −5/2

Multiply the 3rd row by −2/5 : 1 0 1

0 1 3/20 0 1

We already know that the matrix A is invertible. Lets proceedtowards reduced row echelon form .




Add −3 times the 2nd row to the 3rd row:

1 0 10 1 3/20 0 −5/2


0 1 3/20 0 1






Multiply the 3rd row by −2/5 : 1 0 10 1 3/20 0 1






Multiply the 3rd row by −2/5 :

1 0 10 1 3/20 0 1







0 1 3/20 0 1







0 1 3/20 0 1

We already know t

hat the matrix A is invertible. Lets proceedtowards reduced row echelon form .






0 1 3/20 0 1

We already know that the matrix A

is invertible. Lets proceedtowards reduced row echelon form .






0 1 3/20 0 1

We already know that the matrix A is invertible.

Lets proceedtowards reduced row echelon form .






0 1 3/20 0 1

We already know that the matrix A is invertible. Lets proceed

towards reduced row echelon form .






0 1 3/20 0 1

We already know that the matrix A is invertible. Lets proceedtowards reduced

row echelon form .






0 1 3/20 0 1





Add −3/2 times the 3rd row to the 2nd row: 1 0 10 1 00 0 1

Add −1 times the 3rd row to the 1st row: 1 0 0

0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:




Add −3/2 times the 3rd row to the 2nd row:

1 0 10 1 00 0 1


0 1 00 0 1






Add −1 times the 3rd row to the 1st row: 1 0 00 1 00 0 1






Add −1 times the 3rd row to the 1st row:

1 0 00 1 00 0 1







0 1 00 0 1







0 1 00 0 1

To obtain A−1,

in this case, we need to apply the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case,

we need to apply the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need

to apply the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need to apply

the followingsequence of elementary row operations ( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of

elementary row operations ( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of elementary row operations

( same applied to A before)to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of elementary row operations ( same applied to A before)

to the identity matrix:






0 1 00 0 1

To obtain A−1, in this case, we need to apply the followingsequence of elementary row operations ( same applied to A before)to the

identity matrix:






0 1 00 0 1





1) Interchange the 1st row with the 2nd row.

2) Add −3 times the 1st row to the 2nd row.

3) Add 2 times the 1st row to the 3rd row.

4) Multiply the 2nd row by −1/2 :

5) Add −3 times the 2nd row to the 3rd row.

6) Multiply the 3rd row by −2/5

7) Add −3/2 times the 3rd row to the 2nd row.

8) Add −1 times the 3rd row to the 1st row.




























































































A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way

to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute

the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1

is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices

A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I

into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one

3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix

(a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I )

(called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called

TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix),

and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations

to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix,

until A is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A

is transformed into I and the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and

the identity matrix Iwill be automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be

automatically transformed into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




A convenient way to compute the inverse matrix A−1 is to writethe matrices A and I into one 3× 6 matrix (a|I ) (called TheAugmented Matrix), and apply elementary row operations to thisnew matrix, until A is transformed into I and the identity matrix Iwill be automatically transformed

into A−1

A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1





A =

3 −2 01 0 1−2 3 0

I =

1 0 00 1 00 0 1

(A|I ) =

3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1

Interchange the 1st row with the 2nd row: 1 0 1 0 1 0

3 −2 0 1 0 0−2 3 0 0 0 1

Add −3 times the 1st row to the 2nd row: 1 0 1 0 1 0

0 −2 −3 1 −3 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1

Interchange the 1st row with the 2nd row: 1 0 1 0 1 03 −2 0 1 0 0−2 3 0 0 0 1


0 −2 −3 1 −3 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1

Interchange the 1st row with the 2nd row:

1 0 1 0 1 03 −2 0 1 0 0−2 3 0 0 0 1


0 −2 −3 1 −3 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1


3 −2 0 1 0 0−2 3 0 0 0 1

Add −3 times the 1st row to the 2nd row: 1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1


3 −2 0 1 0 0−2 3 0 0 0 1

Add −3 times the 1st row to the 2nd row:

1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1




3 −2 0 1 0 01 0 1 0 1 0−2 3 0 0 0 1


3 −2 0 1 0 0−2 3 0 0 0 1


0 −2 −3 1 −3 0−2 3 0 0 0 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1

Add 2 times the 1st row to the 3rd row. 1 0 1 0 1 0

0 −2 −3 1 −3 00 3 2 0 2 1

Multiply the 2nd row by −1/2 : 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1

Add 2 times the 1st row to the 3rd row. 1 0 1 0 1 00 −2 −3 1 −3 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1

Add 2 times the 1st row to the 3rd row.

1 0 1 0 1 00 −2 −3 1 −3 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1


0 −2 −3 1 −3 00 3 2 0 2 1

Multiply the 2nd row by −1/2 : 1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1


0 −2 −3 1 −3 00 3 2 0 2 1

Multiply the 2nd row by −1/2 :

1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 −2 −3 1 −3 0−2 3 0 0 0 1


0 −2 −3 1 −3 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 3 2 0 2 1




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1

Add −3 times the 2nd row to the 3rd row. 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0

Multiply the 3rd row by −2/5 1 0 1 0 1 0

0 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1

Add −3 times the 2nd row to the 3rd row. 1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0


0 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1

Add −3 times the 2nd row to the 3rd row.

1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0


0 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0

Multiply the 3rd row by −2/5 1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0

Multiply the 3rd row by −2/5

1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 3 2 0 2 1


0 1 3/2 −1/2 3/2 00 0 5/2 −3/5 1 0


0 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5

Add −3/2 times the 3rd row to the 2nd row. 1 0 1 0 1 0

0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

Add −1 times the 3rd row to the 1st row. 1 0 0 3/5 0 2/5

0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5

Add −3/2 times the 3rd row to the 2nd row. 1 0 1 0 1 00 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5

Add −3/2 times the 3rd row to the 2nd row.

1 0 1 0 1 00 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

Add −1 times the 3rd row to the 1st row. 1 0 0 3/5 0 2/50 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

Add −1 times the 3rd row to the 1st row.

1 0 0 3/5 0 2/50 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




1 0 1 0 1 00 1 3/2 −1/2 3/2 00 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5


0 1 0 2/5 0 3/50 0 1 −3/5 1 −2/5

= (I |A−1)




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1

3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=

1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 11 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is

3 −2 11 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




Thus

3 −2 11 0 1−2 3 0

−1 3/5 0 2/52/5 0 3/5−3/5 1 −2/5

=1

5

3 0 22 0 3−3 5 −2

That is 3 −2 1

1 0 1−2 3 0

1

5

3 0 22 0 3−3 5 −2

=

1 0 00 1 00 0 1




We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.

Question: Why does it work?

Proposition

Any elementary row operation can be simulated as leftmultiplication by a certain ( elementary ) matrix




We already saw

( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5)

that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way

to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to compute

the inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix

A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1

is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices

A and I into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I

into onematrix (A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix

(A|I ) and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I )

and apply elementary row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary

row operations to this newmatrix.


Proposition





We already saw ( Theorem 3.5) that a convenient way to computethe inverse matrix A−1 is to merge the matrices A and I into onematrix (A|I ) and apply elementary row operations

to this newmatrix.


Proposition







Proposition







Proposition







Proposition







Proposition

Any

elementary row operation can be simulated as leftmultiplication by a certain ( elementary ) matrix






Proposition

Any elementary row operation

can be simulated as leftmultiplication by a certain ( elementary ) matrix






Proposition

Any elementary row operation can be simulated

as leftmultiplication by a certain ( elementary ) matrix






Proposition

Any elementary row operation can be simulated as leftmultiplication by

a certain ( elementary ) matrix






Proposition

Any elementary row operation can be simulated as leftmultiplication by a certain

( elementary ) matrix






Proposition





Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i





Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i





Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix

EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA

can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained

from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A,

multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying

the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row

byr . (The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr .

(The matrix AE can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE

can be obtained from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained

from A, multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A,

multiplying the ithcolumn by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ith

column by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i

The matrix EA can be obtained from A, multiplying the ith row byr . (The matrix AE can be obtained from A, multiplying the ithcolumn

by r)




Elementary matrices

1)

E =

1 0. . .

1r

1

0. . .

1

row i





Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j





Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j





Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix

EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA

can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j


from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j


adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times

the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith row

to the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row.

(The matrix AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix

AE can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE

can be obtained from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained

from A, adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A,

adding rtimes the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes

the jth column to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column

to the ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j

The matrix EA can be obtained from A, adding r times the ith rowto the jth row. (The matrix AE can be obtained from A, adding rtimes the jth column to the

ith column)




Elementary matrices

2)

E =

1 0...

. . .... · · · 1...

.... . .

0 · · · r · · · 1...

......

. . .

0 · · · 0 · · · 0 · · · 1

row i

row j





Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j





Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j





Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix

EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA

can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j


from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j


interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging

the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith row

with the jth row. (The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row.

(The matrix AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix

AE can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE

can be obtained from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained

from A,interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,

interchanging the ith column with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j

The matrix EA can be obtained from A, interchanging the ith rowwith the jth row. (The matrix AE can be obtained from A,interchanging the ith column

with the jth column)




Elementary matrices

3)

E =

1 0. . .

0 · · · 1...

. . ....

1 · · · 0. . .

0 1

row i

row j





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.

Applying the same sequence of operations to the identity matrix,we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1




Thus,

assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that

a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A

can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted

to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix

by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of

elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations.

ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A

where E1,E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,

E2, · · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2,

· · · ,Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,

Ek−1,Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,

Ek are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek

are elementarymatrices simulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Thus, assume that a square matrix A can be converted to theidentity matrix by a sequence of elementary row operations. ThenEkEk−1 · · ·E2E1A where E1,E2, · · · ,Ek−1,Ek are elementarymatrices s

imulating those operations.


B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1







B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1






Applying

the same sequence of operations to the identity matrix,we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1






Applying the same sequence

of operations to the identity matrix,we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1






Applying the same sequence of operations

to the identity matrix,we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1






Applying the same sequence of operations to the identity matrix,

we obtain the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1






Applying the same sequence of operations to the identity matrix,we obtain

the matrix

B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1







B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1







B = EkEk−1 · · ·E2E1I =

EkEk−1 · · ·E2E1







B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus,

BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I .

Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover,

B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible

since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matrices

are invertible. It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible.

It follows that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows

that B−1(BA) = B−1I , then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I ,

then A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then

A = B−1, soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1

Thus, BA = I . Moreover, B is invertible since elementary matricesare invertible. It follows that B−1(BA) = B−1I , then A = B−1,

soB = A−1






B = EkEk−1 · · ·E2E1I = EkEk−1 · · ·E2E1





Transpose of a matrix

Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given

a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A

the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A,

denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT ,

is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix

whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows

are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A

(and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns

are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A )

That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is,

if A = (aij), then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij),

then AT = (bij) where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij)

where bij = aji . Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji .

Thus,for instance (

1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,

for instance (1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)





Given a matrix A the transpose of A, denoted by AT , is thematrix whose rows are columns of A (and whose columns are rowsof A ) That is, if A = (aij), then AT = (bij) where bij = aji . Thus,for instance

(1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

145

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=

(1 4 5

) 1 2

2 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)






1 2 34 5 6

)T

=

1 42 53 6

1

45

T

=(

1 4 5)

1 22 05

T

=

(1 22 0

)Dr. Marco A Roque Sol Linear Algebra. Session 3



Properties of transposes

(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1





(AT )T = A

(A + B)T = AT + BT

(rA)T = rAT

(AB)T = BTAT

(A1A2 · · ·Ak)T = ATk A

Tk−1 · · ·AT

2 AT1

(A−1)T = (AT )−1




Definition

A square matrix A is said to be symmetric if A = AT

For example, any diagonal matrix is symmetric.

Proposition

For any square matrix A the matrices B = AAT and C = A + AT ,are symmetric.




Definition



Proposition





Definition



Proposition





Definition

A square matrix

A is said to be symmetric if A = AT


Proposition





Definition

A square matrix A

is said to be symmetric if A = AT


Proposition





Definition

A square matrix A is said to be

symmetric if A = AT


Proposition





Definition

A square matrix A is said to be symmetric

if A = AT


Proposition





Definition



Proposition





Definition


For example,

any diagonal matrix is symmetric.

Proposition





Definition


For example, any diagonal matrix

is symmetric.

Proposition





Definition



Proposition





Definition



Proposition





Definition



Proposition

For any

square matrix A the matrices B = AAT and C = A + AT ,are symmetric.




Definition



Proposition

For any square matrix A

the matrices B = AAT and C = A + AT ,are symmetric.




Definition



Proposition

For any square matrix A the matrices B = AAT and

C = A + AT ,are symmetric.




Definition



Proposition




Determinants I

The general definition of the determinant is quite complicated asthere is no simple explicit formula

Approach 1 (Axiomatic): We formulate properties that thedeterminant should have.

Approach 2 (Inductive): The determinant of an n × n matrix isdefined in terms of determinants of certain (n − 1)× (n − 1)matrices.

Approach 3 (Original): An explicit (but very complicated)formula is provided.



Determinants I

The general definition

of the determinant is quite complicated asthere is no simple explicit formula






Determinants I

The general definition of the determinant

is quite complicated asthere is no simple explicit formula






Determinants I

The general definition of the determinant is quite complicated

asthere is no simple explicit formula






Determinants I

The general definition of the determinant is quite complicated asthere is no

simple explicit formula






Determinants I







Determinants I


Approach 1 (Axiomatic):

We formulate properties that thedeterminant should have.





Determinants I


Approach 1 (Axiomatic): We formulate

properties that thedeterminant should have.





Determinants I


Approach 1 (Axiomatic): We formulate properties that

thedeterminant should have.





Determinants I


Approach 1 (Axiomatic): We formulate properties that thedeterminant

should have.





Determinants I







Determinants I



Approach 2 (Inductive):

The determinant of an n × n matrix isdefined in terms of determinants of certain (n − 1)× (n − 1)matrices.




Determinants I



Approach 2 (Inductive): The determinant of

an n × n matrix isdefined in terms of determinants of certain (n − 1)× (n − 1)matrices.




Determinants I



Approach 2 (Inductive): The determinant of an n × n matrix

isdefined in terms of determinants of certain (n − 1)× (n − 1)matrices.




Determinants I



Approach 2 (Inductive): The determinant of an n × n matrix isdefined

in terms of determinants of certain (n − 1)× (n − 1)matrices.




Determinants I



Approach 2 (Inductive): The determinant of an n × n matrix isdefined in terms of

determinants of certain (n − 1)× (n − 1)matrices.




Determinants I



Approach 2 (Inductive): The determinant of an n × n matrix isdefined in terms of determinants of

certain (n − 1)× (n − 1)matrices.




Determinants I







Determinants I




Approach 3 (Original):

An explicit (but very complicated)formula is provided.



Determinants I




Approach 3 (Original): An explicit

(but very complicated)formula is provided.



Determinants I




Approach 3 (Original): An explicit (but very complicated)

formula is provided.



Determinants I




Approach 3 (Original): An explicit (but very complicated)formula

is provided.



Determinants I







Determinants I

Approach 1 (Axiomatic)

Mn×n(R): The set of n × n matrices with real entries.

AXIOMS

There exists a unique function det :Mn×n(R) −→ R (called thedeterminant) with the following properties

A1 If a row of a matrix is multiplied by a scalar r , the determinantis also multiplied by r

A2 If we add a row of a matrix multiplied by a scalar to anotherrow, the determinant remains the same

A3 If we interchange two rows of a matrix, the determinantchanges its sign;

A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I


Mn×n(R):

The set of n × n matrices with real entries.

AXIOMS





A4: det(I ) = 1



Determinants I


Mn×n(R): The set of

n × n matrices with real entries.

AXIOMS





A4: det(I ) = 1



Determinants I


Mn×n(R): The set of n × n matrices

with real entries.

AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS

There exists

a unique function det :Mn×n(R) −→ R (called thedeterminant) with the following properties




A4: det(I ) = 1



Determinants I



AXIOMS

There exists a unique function det :Mn×n(R) −→ R

(called thedeterminant) with the following properties




A4: det(I ) = 1



Determinants I



AXIOMS

There exists a unique function det :Mn×n(R) −→ R (called thedeterminant)

with the following properties




A4: det(I ) = 1



Determinants I



AXIOMS

There exists a unique function det :Mn×n(R) −→ R (called thedeterminant) with the following

properties




A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS


A1 If a row of

a matrix is multiplied by a scalar r , the determinantis also multiplied by r



A4: det(I ) = 1



Determinants I



AXIOMS


A1 If a row of a matrix

is multiplied by a scalar r , the determinantis also multiplied by r



A4: det(I ) = 1



Determinants I



AXIOMS


A1 If a row of a matrix is multiplied by

a scalar r , the determinantis also multiplied by r



A4: det(I ) = 1



Determinants I



AXIOMS


A1 If a row of a matrix is multiplied by a scalar r ,

the determinantis also multiplied by r



A4: det(I ) = 1



Determinants I



AXIOMS


A1 If a row of a matrix is multiplied by a scalar r , the determinant

is also multiplied by r



A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add

a row of a matrix multiplied by a scalar to anotherrow, the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of

a matrix multiplied by a scalar to anotherrow, the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of a matrix

multiplied by a scalar to anotherrow, the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of a matrix multiplied by

a scalar to anotherrow, the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of a matrix multiplied by a scalar

to anotherrow, the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of a matrix multiplied by a scalar to anotherrow,

the determinant remains the same


A4: det(I ) = 1



Determinants I



AXIOMS



A2 If we add a row of a matrix multiplied by a scalar to anotherrow, the determinant remains

the same


A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS




A3 If we interchange

two rows of a matrix, the determinantchanges its sign;

A4: det(I ) = 1



Determinants I



AXIOMS




A3 If we interchange two rows

of a matrix, the determinantchanges its sign;

A4: det(I ) = 1



Determinants I



AXIOMS




A3 If we interchange two rows of a matrix,

the determinantchanges its sign;

A4: det(I ) = 1



Determinants I



AXIOMS




A3 If we interchange two rows of a matrix, the determinant

changes its sign;

A4: det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I



AXIOMS





A4:

det(I ) = 1



Determinants I



AXIOMS





A4: det(I ) = 1



Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7

det(B) = 0 whenever the matrix B has a zero row

Theorem 3.8

det(A) = 0 if and only if the matrix is not invertible.



Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A

is a square matrix and B is obtained from A applyingelementary row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and

B is obtained from A applyingelementary row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A

applyingelementary row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary

row operations. Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary row operations.

Then det(A) = 0 if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary row operations. Then det(A) = 0

if and only ifdet(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6

Suppose A is a square matrix and B is obtained from A applyingelementary row operations. Then det(A) = 0 if and only if

det(B) = 0.

Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7

det(B) = 0

whenever the matrix B has a zero row

Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7

det(B) = 0 whenever

the matrix B has a zero row

Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7

det(B) = 0 whenever the matrix B

has a zero row

Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8

det(A) = 0

if and only if the matrix is not invertible.



Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8

det(A) = 0 if and only if

the matrix is not invertible.



Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8

det(A) = 0 if and only if the matrix

is not invertible.



Determinants I

Theorem 3.6


Theorem 3.7


Theorem 3.8




Determinants I

Idea of the proof: Let B be the reduced row echelon form of A. IfA is invertible then B = I ; otherwise B has a zero row .

Remark. The same argument proves that properties (A1)(A4) areenough to evaluate any determinant.

Theorem 3.9

If a matrix A has two proportional rows then det(A) = 0



Determinants I

Idea of the proof:

Let B be the reduced row echelon form of A. IfA is invertible then B = I ; otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be

the reduced row echelon form of A. IfA is invertible then B = I ; otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be the reduced row

echelon form of A. IfA is invertible then B = I ; otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be the reduced row echelon form of A.

IfA is invertible then B = I ; otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be the reduced row echelon form of A. IfA is invertible

then B = I ; otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be the reduced row echelon form of A. IfA is invertible then B = I ;

otherwise B has a zero row .


Theorem 3.9




Determinants I

Idea of the proof: Let B be the reduced row echelon form of A. IfA is invertible then B = I ; otherwise B

has a zero row .


Theorem 3.9




Determinants I



Theorem 3.9




Determinants I


Remark. The same argument

proves that properties (A1)(A4) areenough to evaluate any determinant.

Theorem 3.9




Determinants I


Remark. The same argument proves that properties

(A1)(A4) areenough to evaluate any determinant.

Theorem 3.9




Determinants I


Remark. The same argument proves that properties (A1)(A4)

areenough to evaluate any determinant.

Theorem 3.9




Determinants I


Remark. The same argument proves that properties (A1)(A4) areenough

to evaluate any determinant.

Theorem 3.9




Determinants I


Remark. The same argument proves that properties (A1)(A4) areenough to evaluate

any determinant.

Theorem 3.9




Determinants I



Theorem 3.9




Determinants I



Theorem 3.9




Determinants I



Theorem 3.9

If a matrix A

has two proportional rows then det(A) = 0



Determinants I



Theorem 3.9

If a matrix A has two

proportional rows then det(A) = 0



Determinants I



Theorem 3.9

If a matrix A has two proportional rows then

det(A) = 0



Determinants I



Theorem 3.9




Determinants I

Row echelon form of a square matrix

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0

det(A) = 0



Determinants I


� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� � ∗ ∗ ∗� ∗ ∗

� �

det(A) 6= 0 det(A) = 0



Determinants I

As an example, we have the matrix

A =

3 −2 01 0 1−2 3 0

Earlier we have transformed the matrix A into the identity matrixusing elementary row operations.

These included two row multiplications, by −0.5 and by −0.4, andone row exchange.

It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)

Hence det(A) = −5det(I ) = − 5



Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0

Earlier

we have transformed the matrix A into the identity matrixusing elementary row operations.


It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0

Earlier we have transformed

the matrix A into the identity matrixusing elementary row operations.


It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0

Earlier we have transformed the matrix A

into the identity matrixusing elementary row operations.


It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0

Earlier we have transformed the matrix A into the identity matrix

using elementary row operations.


It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0

Earlier we have transformed the matrix A into the identity matrixusing elementary

row operations.


It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0


These included

two row multiplications, by −0.5 and by −0.4, andone row exchange.

It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0


These included two row multiplications,

by −0.5 and by −0.4, andone row exchange.

It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0


These included two row multiplications, by −0.5 and

by −0.4, andone row exchange.

It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0


These included two row multiplications, by −0.5 and by −0.4, and

one row exchange.

It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)




Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)

Hence

det(A) = −5det(I ) = − 5



Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)

Hence det(A) = −5

det(I ) = − 5



Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)

Hence det(A) = −5det(I ) =

− 5



Determinants I


A =

3 −2 01 0 1−2 3 0



It follows that

det(I ) = −(−0.5)(−0.4)det(A) = (−0.2)det(A)

Hence det(A) = −5det(I ) = − 5Dr. Marco A Roque Sol Linear Algebra. Session 3


Determinants I

Approach 2 (Inductive)

Let’s start by considering a 2× 2 system{ax + by = ecx + dy = f

whose solution is given by

x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc


a bc d

)



Determinants I


Let’s start by considering a 2× 2 system

{ax + by = ecx + dy = f


x =ed − bf

ad − bcy =

af − ce

ad − bc


a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc


a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc


a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where

we can notice that the amount ad − bc playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice

that the amount ad − bc playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount

ad − bc playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc

playing an importantrole in the solution is a combination of the elements of the matrix(

a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole

in the solution is a combination of the elements of the matrix(a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution

is a combination of the elements of the matrix(a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution is a combination

of the elements of the matrix(a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution is a combination of the elements

of the matrix(a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc

where we can notice that the amount ad − bc playing an importantrole in the solution is a combination of the elements of the matrix

(a bc d

)



Determinants I




x =ed − bf

ad − bcy =

af − ce

ad − bc


a bc d

)Dr. Marco A Roque Sol Linear Algebra. Session 3


Determinants I

In a similar way, if we consider a 3× 3 systema11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3

we find, the solutions for the variables have in the denominator thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I

In a similar way,

if we consider a 3× 3 systema11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3


a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I

In a similar way, if we consider

a 3× 3 systema11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3


a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I

In a similar way, if we consider a 3× 3

systema11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3


a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I

In a similar way, if we consider a 3× 3 system

a11x + a12y + a13 = b1a21x + a22y + a23 = b2a31x + a32y + a33 = b3


a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I


we find,

the solutions for the variables have in the denominator thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I


we find, the solutions

for the variables have in the denominator thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I


we find, the solutions for the variables

have in the denominator thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I


we find, the solutions for the variables have in the denominator

thequantity

a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which

also plays an important role in the solution set. Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays

an important role in the solution set. Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role

in the solution set. Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set.

Thus, thiscould be a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus,

thiscould be a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus, thiscould be

a practical starting point of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus, thiscould be a practical starting point

of the concept of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)

which also plays an important role in the solution set. Thus, thiscould be a practical starting point of the concept

of determinant.



Determinants I



a11(a22a33 − a23a32)− a12(a21a33 − a23a31)+

a13(a21a32 − a22a31)




Determinants I

Before introducing the concept of determinant, let’s start with acouple of definitions

Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Before introducing

the concept of determinant, let’s start with acouple of definitions

Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Before introducing the concept of determinant,

let’s start with acouple of definitions

Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Before introducing the concept of determinant, let’s start

with acouple of definitions

Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Before introducing the concept of determinant, let’s start with acouple

of definitions

Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition

Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A

a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix

of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A

is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrix

obtained by specifying k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying

k columns and k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and

k rows of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and k rows

of A and deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and

deleting theother columns and rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices

Definition Given a matrix A a k × k submatrix of A is a matrixobtained by specifying k columns and k rows of A and deleting theother columns and

rows.

1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒

∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒

(2 45 9

)



Determinants I


Submatrices


1 2 3 410 20 30 403 5 7 9

⇒ ∗ 2 ∗ 4∗ ∗ ∗ ∗∗ 5 ∗ 9

⇒ (2 45 9

)



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition

Associated to every n × n matrix A there is a real number calledthe determinant of A denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given

an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n

matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij),

let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij ,

denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote

the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant

ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe

(n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix

obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by

deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand

the jth column of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column

of A. That number obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number

obtained in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained

in this way iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way

iscalled the ij-minor of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors

Given an n× n matrix A = (aij), let Mij , denote the determinant ofthe (n − 1)× (n − 1) submatrix obtained by deleting the ith rowand the jth column of A. That number obtained in this way iscalled the ij-minor

of A

Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition

Associated to every n × n matrix A

there is a real number calledthe determinant of A denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition

Associated to every n × n matrix A there is a real number calledthe determinant of A

denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition

Associated to every n × n matrix A there is a real number calledthe determinant of A denoted by |A| or det(A) and

definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1

A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11

|A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2

A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)

|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ =

a11a22 − a12a21



Determinants I

Minors


Definition


n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21Dr. Marco A Roque Sol Linear Algebra. Session 3


Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3

A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣−

a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+

a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

|A| =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

a11

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+



Determinants I

a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12 =

a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−a32a23a11 − a33a21a12

=

∣∣∣∣∣∣∣∣a11 a12 a13 a11 a12a21 a22 a23 a21 a22a31 a32 a33 a31 a32

− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

a12a23a31 +

a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12 =


=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

a12a23a31 + a13a21a32 −

a31a22a13 − a32a23a11 − a33a21a12 =


=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

a12a23a31 + a13a21a32 − a31a22a13 −

a32a23a11 − a33a21a12 =


=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 −

a33a21a12 =


=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I


a11a22a33 + a12a23a31+

a13a21a32 − a31a22a13−a32a23a11 − a33a21a12

=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I


a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−

a32a23a11 − a33a21a12

=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I


a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−a32a23a11 −

a33a21a12

=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣

Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣Now,

for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4,

if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j

be the corresponding minors to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j be the corresponding minors

to the firstrow, then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow,

then we have

|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I



=


− − − + + +


|A| =n∑

j=1

(−1)1+ja1jM1j



Determinants I

Theorem 3.10

For any 1 ≤ k ,m ≤ n we have that

det(A) =n∑

j=1

(−1)k+jakjMij

(expansion by the k − th row )

det(A) =n∑

i=1

(−1)i+maimMij

(expansion by the m − th row )



Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10

For any

1 ≤ k ,m ≤ n we have that

det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10

For any 1 ≤ k ,m ≤ n

we have that

det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.10


det(A) =n∑

j=1

(−1)k+jakjMij


det(A) =n∑

i=1

(−1)i+maimMij




Determinants I

Theorem 3.11

Given an n × n matrix A, if B is an n × n matrix obtained from Aby

1) Additive law for rows

Suppose that matrices X ,Y ,Z are identical except for the ith row.The the ith row of Z is the sum of the ith rows of X and Y then

det(Z ) = det(X ) + det(Y )

a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11

Given an n × n matrix A,

if B is an n × n matrix obtained from Aby




a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

Theorem 3.11





a1 + a′1 a2 + a′2 a3 + a′3b1 b2 b3c1 c2 c3

=

a1 a2 a3b1 b2 b3c1 c2 c3

+

a′1 a′2 a′3b1 b2 b3c1 c2 c3



Determinants I

2) Adding a multiple of the ith row to the jth row then |B| = |A|∣∣∣∣∣∣a1 + rb1 a2 + rb2 a3 + rb3

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then |B| = −|A|3) Interchanging two rows, then |B| = −|A|4) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I

2) Adding a multiple of the ith row to the jth row then

|B| = |A|∣∣∣∣∣∣a1 + rb1 a2 + rb2 a3 + rb3

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I

2) Adding a multiple of the ith row to the jth row then |B| = |A|

∣∣∣∣∣∣a1 + rb1 a2 + rb2 a3 + rb3

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣

2) Interchanging two rows, then |B| = −|A|3) Interchanging two rows, then |B| = −|A|4) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then

|B| = −|A|3) Interchanging two rows, then |B| = −|A|4) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then |B| = −|A|3) Interchanging two rows, then

|B| = −|A|4) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then |B| = −|A|3) Interchanging two rows, then |B| = −|A|

4) Multiplying a row by a nonzero scalar α then |B| = α|A|



Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣2) Interchanging two rows, then |B| = −|A|3) Interchanging two rows, then |B| = −|A|4) Multiplying a row by a nonzero scalar α then

|B| = α|A|



Determinants I


b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣+

∣∣∣∣∣∣rb1 rb2 rb3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3




Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|

3) If A has a row (column) of zeros, then |A| = 0

4) If A has a two identical rows (columns), then |A| = 0



Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence,

for every property of determinants involvingrows of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property

of determinants involvingrows of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants

involvingrows of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows

of a matrix there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix

there is an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix there is

an analogous property involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix there is an analogous property

involving columnsof a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|

( As a consequence, for every property of determinants involvingrows of a matrix there is an analogous property involving columns

of a matrix. )

2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|

3) If A has a row (column) of zeros, then

|A| = 0




Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|


4) If A has a two identical rows (columns), then

|A| = 0



Determinants I

Theorem 3.12

1) |AT | = |A|


2) |AB| = |A||B|





Determinants I

5) If two rows (columns) of A are proportional, then |A| = 0

6) If A is an upper (lower) triangular matrix, then|A| = a11a22a33 · · · ann

7) If A is an invertible matrix, then |A−1| = |A|−1

8) If A is an n × n matrix and r ∈ R, then |rA| = rn|A|



Determinants I

5) If two rows (columns) of A are proportional, then

|A| = 0






Determinants I







Determinants I


6) If A is an upper (lower) triangular matrix, then

|A| = a11a22a33 · · · ann





Determinants I







Determinants I



7) If A is an invertible matrix, then

|A−1| = |A|−1




Determinants I







Determinants I




8) If A is an n × n matrix and r ∈ R, then

|rA| = rn|A|



Determinants I






Linear Algebra. Session 3 - Texas A&M Universityroquesol/Math_304... · Linear Algebra. Session 3 Dr. Marco A Roque Sol 01/30/2018 Dr. Marco A Roque Sol Linear Algebra. Session 3.

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