College Algebra 9th txtbk

SECTION 7–4 Solving Systems of Linear Equations Using Matrix Inverse Methods 475
sequence of steps as in the solution of linear systems by Gauss–Jordan elimination (see
Section 7-2):
A
1 1 1 1 0 0
£ 0 2 1 † 0 1 0§
2 3 0 0 0 1
I
(2)R 1 R 3 S R 3
1 1 1 1 0 0
£ 0 2 1 † 0 1 0§
0 5 2 2 0 1
1 1 1 1 0 0
£ 0 1 1 1
2 † 0 2 0 §
0 5 2 2 0 1
1
1 0 2 1 2 0
£ 0 1 1 1
2 † 0 2 0 §
1
0 0 2 2 5 2 1
1
2 1
1
2 0
1 0
£ 0 1 1 1
2 † 0 2 0 §
0 0 1 4 5 2
1
1
2R 2 S R 2
R 2 R 1 S R 1
(5)R 2 R 3 S R 3
2R 3 S R 3
( 1 2)R 3 R 1 S R 1
1
2R 3 R 2 S R 2
0
1 0 0 3 3 1
£ 0 1 0 † 2 2 1§ [I B]
0 0 1 4 5 2
We suspect that matrix B is actually A 1 , but we should check.
CHECK Because the definition of matrix inverse requires that
A 1 A I and AA 1 I
(3)
it appears that we must compute both A 1 A and to check our work. However, it can be
shown that if one of the equations in (3) is satisfied, then the other is also satisfied. So, for
checking purposes it’s enough to compute either A 1 A or AA 1 —we don’t need to do both.
AA 1
3 3 1 1 1 1 1 0 0
A 1 A £ 2 2 1§£
0 2 1 § £ 0 1 0§ I
4 5 2 2 3 0 0 0 1
MATCHED PROBLEM 1
3 1 1
Let A £ 1 1 0§
1 0 1
(A) Form the augmented matrix
[A | I ].
(B) Use row operations to transform
[A | I ]
into
[I | B].
(C) Verify by multiplication that B A 1 .
The procedure used in Example 1 can be used to find the inverse of any square matrix
if the inverse exists, and will also indicate when the inverse does not exist. These ideas are
summarized in Theorem 1.