1 1. 4 Linear Equations in Linear Algebra THE MATRIX EQUATION © 2016 Pearson Education, Inc. x b A
1 1.4 Linear Equations in Linear Algebra THE MATRIX EQUATION © 2016 Pearson Education, Inc.
Jan 12, 2016
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1
1.4
Linear Equationsin Linear Algebra
THE MATRIX EQUATION x bA
© 2016 Pearson Education, Inc.
Slide 1.4- 2
MATRIX EQUATION Definition: If A is an matrix, with columns a1,
…, an, and if x is in , then the product of A and x, denoted by Ax, is the linear combination of the columns of A using the corresponding entries in x as weights; that is,
.
Note that Ax is defined only if the number of columns of A equals the number of entries in x.
x bA m n
1
2
1 2 1 1 2 2x a a a a a ... an n n
n
x
xA x x x
x
© 2016 Pearson Education, Inc.
Slide 1.4- 3
MATRIX EQUATION
Example 2: For v1, v2, v3 in , write the linear combination as a matrix times a vector.
Solution: Place v1, v2, v3 into the columns of a matrix A and place the weights 3, , and 7 into a vector x.
That is,
.
x bA
1 2 33v 5v 7v
5
1 2 3 1 2 3
3
3v 5v 7v v v v 5 x
7
A
© 2016 Pearson Education, Inc.
Slide 1.4- 4
MATRIX EQUATION
Now, write the system of linear equations as a vector equation involving a linear combination of vectors.
For example, the following system
(1)
is equivalent to
. (2)
x bA
1 2 3
2 3
2 4
5 3 1
x x x
x x
1 2 3
1 2 1 4
0 5 3 1x x x
© 2016 Pearson Education, Inc.
Slide 1.4- 5
MATRIX EQUATION
As in the example, the linear combination on the left side is a matrix times a vector, so that (2) becomes
. (3)
Equation (3) has the form . Such an equation
is called a matrix equation, to distinguish it from a vector equation such as shown in (2).
x bA
1
2
3
1 2 1 4
0 5 3 1
x
x
x
x bA
© 2016 Pearson Education, Inc.
Slide 1.4- 6
MATRIX EQUATION x bA
© 2016 Pearson Education, Inc.
If A is an matrix, with columns a1, …, an, and if b is in , then the matrix equation
Ax = bhas the same solution set as the vector equation
which, in turn, has the same solution set as the system of linear equations whose augmented matrix is
[ b]
THEOREM 3
Slide 1.4- 7
EXISTENCE OF SOLUTIONS The equation has a solution if and only if b
is a linear combination of the columns of A. x bA
© 2016 Pearson Education, Inc.
Let A be an matrix. Then the following statements are logically equivalent. That is, for a particular A, either they are all true statements or they are all false.
a. For each b in , the equation Ax=b has a solution.
b. Each b in is a linear combination of the columns of A.
c. The columns of A span .d. A has a pivot position in every row.
THEOREM 4
Slide 1.4- 8
COMPUTATION OF Ax
Example 4: Compute Ax, where
and .
Solution: From the definition,
2 3 4
1 5 3
6 2 8
A
1
2
3
x
x
x
x
1
2 1 2 3
3
2 3 4 2 3 4
1 5 3 1 5 3
6 2 8 6 2 8
x
x x x x
x
© 2016 Pearson Education, Inc.
Slide 1.4- 9
COMPUTATION OF Ax
(1)
. The first entry in the product Ax is a sum of products
(sometimes called a dot product), using the first row of A and the entries in x.
1 2 3
1 2 3
1 2 3
2 3 4
5 3
6 2 8
x x x
x x x
x x x
1 2 3
1 2 3
1 2 3
2 3 4
5 3
6 2 8
x x x
x x x
x x x
© 2016 Pearson Education, Inc.
Slide 1.4- 10
COMPUTATION OF Ax
That is, .
Similarly, the second entry in Ax can be calculated by multiplying the entries in the second row of A by the corresponding entries in x and then summing the resulting products.
1 1 2 3
2
3
2 3 4 2 3 4x x x x
x
x
1
2 1 2 3
3
1 5 3 5 3
x
x x x x
x
© 2016 Pearson Education, Inc.
Slide 1.4- 11
ROW-VECTOR RULE FOR COMPUTING Ax
Likewise, the third entry in Ax can be calculated from the third row of A and the entries in x.
If the product Ax is defined, then the ith entry in Ax is the sum of the products of corresponding entries from row i of A and from the vector x.
The matrix with 1’s on the diagonal and 0’s elsewhere is called an identity matrix and is denoted by I.
For example, is an identity matrix.
1 0 0
0 1 0
0 0 1
© 2016 Pearson Education, Inc.
Slide 1.4- 12
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax
Theorem 5: If A is an matrix, u and v are vectors in , and c is a scalar, then
a.
b. . Proof: For simplicity, take , ,
and u, v in . For let ui and vi be the ith entries in u and
v, respectively.
m n
(u v) u v;A A A ( u) ( u)A c c A
3n 1 2 3
a a aA
1,2,3,i
© 2016 Pearson Education, Inc.
Slide 1.4- 13
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax
© 2016 Pearson Education, Inc.
If A is an matrix, u and v are vectors in, and c is a scalar, then
a. A(u + v) = Au + Av;b. A(cu) = c(Au).
THEOREM 5
Proof: For simplicity, take n = 3, A = [a1 a2 a3], and u, v in .
For i = 1, 2, 3, let ui and vi be the ith entries in u and v, respectively.
Slide 1.4- 14
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax To prove statement (a), compute as a linear
combination of the columns of A using the entries in
as weights.
(u v)A
u v
1 1
1 2 3 2 2
3 3
(u v) a a a
u v
A u v
u v
1 1 1 2 2 2 3 3 3( )a ( )a ( )au v u v u v Entries in u v
Columns of A
1 1 2 2 3 3 1 1 2 2 3 3( a a a ) ( a a a )u u u v v v
u vA A © 2016 Pearson Education, Inc.
Slide 1.4- 15
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax
To prove statement (b), compute as a linear combination of the columns of A using the entries in cu as weights.
( u)A c
1
1 2 3 2 1 1 2 2 3 3
3
( u) a a a ( )a ( )a ( )a
cu
A c cu cu cu cu
cu
1 1 2 2 3 3( a ) ( a ) ( a )c u c u c u
1 1 2 2 3 3( a a a )c u u u ( u)c A
© 2016 Pearson Education, Inc.
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