Page 251 - Advanced engineering mathematics

7.7 Matrix Inverses 231

Reduce this by multiplying row one by −1/3 and then adding −4 times row one to row two
to get
.
⎛ ⎞
1 −7 . . −1/3 0
.
⎝ ⎠ .
0 0 . . 4/3 1
The left two columns, which form A R , do not equal I 2 ,so A is singular and has no inverse.
We will illustrate the use of a matrix inverse to solve a nonhomogeneous system.

EXAMPLE 7.30
We will solve the system
2x 1 − x 2 + 3x 3 = 4

x 1 + 9x 2 − 2x 3 =−8
4x 1 − 8x 2 + 11x 3 = 15.
The matrix of coefficients is
⎛ ⎞
2 −1 3
A = 1 9 −2 ⎠ .
⎝
4 −8 11
A routine reduction yields
.
⎛ ⎞
100 . . 83/53 −13/53 −25/53
. ⎜ ⎟
. ⎜ . . ⎟
[A.I 3 ] R = ⎜ 010 . −19/53 10/53 7/53 ⎟.
.
⎝ ⎠
001 . . −44/53 12/53 19/53
The first three columns are I 3 , hence A is nonsingular and the system has a unique solution. The
last three columns of the reduced augmented matrix give us
⎛ ⎞
83 −13 −25
1
−1
A = ⎝ −19 10 7 ⎠ .
53
−44 12 19
−1
The unique solution of the system is A B:
⎛ ⎞⎛ ⎞ ⎛ ⎞
83 −13 −25 4 61/53
1
−1 ⎝ −19 .
X = A B = 10 7 ⎠⎝ −8 ⎠ = −51/53 ⎠
⎝
53
−44 12 19 15 13/53
SECTION 7.7 PROBLEMS

In each of Problems 1 through 10, find the inverse of the −5 2
3.
matrix or show that the matrix is singular. 1 2
−1 0

4.
−1 2
1. 4 4
2 1
6 2

5.
12 3 3 3
2.
4 1
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October 14, 2010 14:23 THM/NEIL Page-231 27410_07_ch07_p187-246