84 CHAPTER 4. MATRICES

Checking the answer is easy. Just multiply the matrices and see if it works. 1 0 11 2 11 1 2

 0 2 2

1 2 01 1 1

=

 1 0 00 1 00 0 1

All arithmetic is done in Z3. Always check your answer because if you are like some of us,you will usually have made a mistake.

Example 4.2.5 Let A =

 6 −1 2−1 2 −12 −1 1

. Find A−1 in Q.

Set up the augmented matrix (A|I) 6 −1 2 1 0 0−1 2 −1 0 1 02 −1 1 0 0 1

Now find row reduced echelon form 1 0 0 1 −1 −3

0 1 0 −1 2 40 0 1 −3 4 11

Thus the inverse is  1 −1 −3

−1 2 4−3 4 11



Example 4.2.6 Let A =

 1 2 21 0 22 2 4

. Find A−1 in Q.

This time there is no inverse because the columns are not linearly independent. Thiscan be seen by solving the equation 1 2 2

1 0 22 2 4

 x

yz

=

 000

and finding that there is a nonzero solution which is equivalent to the columns being adependent set. Thus, by Theorem 4.2.3, there is no inverse.

Example 4.2.7 Consider the matrix  1 1 00 1 00 0 5

Find its inverse in arithmetic of Q and then find its inverse in Z5.