This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Solutions to quiz # 4 (September 30)
1. Computing the inverse, we obtain 1
2
3 1
0
−→
0 0
1
−1
01
1a
01 100
1 −→ 0
010
001
0  1 00
10 −→ 0 1
 −2 1 0
 −5 1 1
00 1
a+2
4−a 0
1
−1
0
0
1 0
0
1 6
−1
−1
5a − 2 1 − a −a .
−5
1
1 1
1
a
 −2
1 − a  −3 0
1
0 Answer: The inverse matrix is 6 5a − 2
−5 −1
1−a
1 −1
−a .
1 2. Solving for X , we obtain
ABXC = D =⇒ A−1 (ABXC )C −1 = A−1 DC −1
=⇒ A−1 A B X C C −1 = A−1 DC −1 =⇒ IBXI = A−1 DC −1
=⇒BX = A−1 DC −1 =⇒ B −1 (BX ) = B −1 A−1 DC −1
=⇒ B −1 B X = B −1 A−1 DC −1 =⇒ IX = B −1 A−1 DC −1
=⇒X = B −1 A−1 DC −1 .
Answer:
X = B −1 A−1 DC −1 . 1 ...
View
Full
Document
This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra

Click to edit the document details