120 CHAPTER 10

SOLUTION: We have

7811

4312

⎡

⎢

⎣

85

39

11 2

⎤

⎥

⎦

=

201 129

173 71

and

⎡

⎢

⎣

85

39

11 2

⎤

⎥

⎦

7811

4312

=

⎡

⎢

⎣

76 79 148

57 51 141

85 94 145

⎤

⎥

⎦

.

The process by which we have deﬁned matrix multiplication is

typically mathematical. First we deﬁne and understand a special

case: a row times a column, that is, a 1-by-n matrix times an n-by-1

matrix, will be a 1-by-1 matrix—a single number—given by the dot

product. From this we build up: next deﬁning any matrix times a

column, and ﬁnally any matrix times any matrix (as long as the

sizes match).

Linear Algebra

One simple use of matrices is to solve systems of Z-equations

where none of the exponents is higher than 1. This is called “linear

algebra.”

2

Here is an example.

Suppose we want to solve the system of Z-equations:

3x − 5y = 2. (10.1)

2x + 3y = 14. (10.2)

This is easy to solve using elementary algebra. And it is also easy

to graph the two lines (which is why it is called linear algebra) and

see where they intersect. But in order to illustrate matrices, let’s

set this up as a matrix problem.

2

This is one reason why “linear” is often used as an adjective in place of “matricial.”

MATRICES 121

We deﬁne the Z-matrix A to be

A =

3 −5

23

and we deﬁne another Z-matrix B to be

B =

2

14

.

Finally, we deﬁne an unknown matrix Z in terms of x and y:

Z =

x

y

.

The value Z is just as unknown as the x and y we are trying to ﬁnd.

Now by practicing your matrix multiplication, you can see that

the system given by (10.1) and (10.2) above is completely equivalent

to the single matrix equation:

AZ = B.

It turns out there is a way to divide by the matrix A and solve

for Z. We will discuss this in more detail in the next chapter. For

now, we just deﬁne two more matrices:

A

−1

=

3

19

5

19

−2

19

3

19

and

I =

10

01

.

Check that A

−1

A = I and IZ = Z.

If AZ = B then A

−1

(AZ) = A

−1

B. In the next chapter we will ﬁnd

out that matrix multiplication is associative, so that A

−1

(AZ) =

(A

−1

A)Z = IZ = Z. We conclude that if AZ = B,then

Z = A

−1

B =

3

19

5

19

−2

19

3

19

2

14

=

76

19

38

19

=

4

2

.

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