left-hand side of the transformed augmented matrix. Consequently, the right-hand
side of equation (10.27) becomes an equation, which gives the inverse matrix. The
following example illustrates this “machine.”
Example 10-23. Find the inverse of the matrix A=
1 0 3
−1 1 2
2 −1 2
.
Solution Append to the matrix Athe identity matrix I to obtain the augmented
matrix
1 0 3
−1 1 2
2 −1 2
∣∣
∣∣
∣∣
1 0 0
0 1 0
0 0 1
. (10 .28)
Now try to select a sequence of elementary row operations with the goal of reducing
the left-hand side of the augmented matrix (10.28) to the identity matrix. Each time
an elementary row operation is applied to to the left-hand side of the augmented
matrix (10.28) be sure to “record” the operation on the right-hand side. To illustrate,
consider the following row operations applied to the augmented matrix (10.28).
- Replace row 2 by adding row 1 to row 2
- Multiply row 1 by (−2) and add the result to row 3. This produces the augmented
matrix
1 0 3
0 1 5
0 − 1 − 4
∣∣
∣∣
∣∣
1 0 0
1 1 0
−2 0 1
.
Next perform the elementary row operation of replacing row 3 by adding row 2
to row 3 to get
1 0 3
0 1 5
0 0 1
∣∣
∣∣
∣∣
1 0 0
1 1 0
−1 1 1
.
Finally, perform the following row operations:
- Multiply row 3 by (−5) and add the result to row 2.
- Multiply row 3 by (−3) and add the result to row 1. The above row operations
produce the desired result of producing the identity matrix on the left-hand side
of the augmented matrix. The final form for the augmented matrix is
1 0 0
0 1 0
0 0 1
∣∣
∣∣
∣∣
4 − 3 − 3
6 − 4 − 5
−1 1 1
and an examination of the right-hand side of the augmented matrix gives the
inverse matrix A−^1 =
4 − 3 − 3
6 − 4 − 5
−1 1 1
.One can readily verify that AA −^1 =I.
