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

**Unformatted text preview: **= y B = 1 A3 = Az = 1 −1
1
A = 1 −1
3
0 −4
z
0
3
0
0
Expanding both det(A) and det (Az ) along the third rows (to take advantage of the 0’s) gives
z= −12
6
det (Az )
=
=
det(A)
−10
5 The reader is encouraged to solve this system for x and y similarly and check the answer.
Our last application of determinants is to develop an alternative method for ﬁnding the inverse of
a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 3
1
2
5
A = 0 −1
2
1
4
We found through a variety of methods that det(A) = −13. To our surprise and delight, its inverse
below has a remarkable number of 13’s in the denominators of its entries. This is no coincidence. 9
2
7
13
13 − 13 8
15 A−1 = − 10 − 13
13
13 2
1
3
− 13
13
13
Recall that to ﬁnd A−1 , we are essentially solving the matrix equation AX = I3 , where X = [xij ]3×3
is a 3 × 3 matrix. Because of how matrix multiplication is deﬁned, the ﬁrst column of I3 is the
product of A with the ﬁrst column of X , the second column of I3...

View
Full
Document