*This preview shows
page 1. Sign up to
view the full content.*

**Unformatted text preview: **s of Equations and Matrices also know that the process by which we ﬁnd A−1 is determined completely by A, and not by the
constants in B . This answers the question as to why we would bother doing row operations on
a super-sized augmented matrix to ﬁnd A−1 instead of an ordinary augmented matrix to solve a
system; by ﬁnding A−1 we have done all of the row operations we ever need to do, once and for all,
since we can quickly solve any equation AX = B using one multiplication, A−1 B . We illustrate
this in our next example. 3
1
2
5
Example 8.4.1. Let A = 0 −1
2
1
4
1. Use row operations to ﬁnd A−1 . Check your answer by ﬁnding A−1 A and AA−1 .
2. Use A−1 to solve the 3x + y + 2 z
−y + 5z
(a) 2x + y + 4 z following systems of equations = 26 3x + y + 2 z = 4
= 39
−y + 5z = 2
(b) = 117
2x + y + 4 z = 5 3x + y + 2 z = 1
−y + 5z = 0
(c) 2x + y + 4 z = 0 Solution.
1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination. 3
1
2 0 −1
5
2
1
4 2
1
1
3
3 0 −1
5
2
1
4 1
2
1
3
3 0 −1
5
8
1
0
3
3 1
2
1
3
3
0
1 −5...

View Full
Document