Unformatted text preview: U so A M 1 U and 211
M L 1.
ii) Changing a 23 to what value would cause A
If a 23 2 we obtain 1 not to exist? 11 1100 1 1 1 1 00 11 1 1 00 01 2010 0 1 2 0 10 01 1 0 10 21 0 0 1 2 201 00 0 211 001 and A is singular. 4 001
5. (8 pts) If we know that 100 0 2 0 101 100 11 110 A 010 1
0 0 1 , solve 0
Ax 1
1 (Hint: first multiply both sides of Ax b on the left by P, do not calculate A !)
0
We have PA LU. Now if Ax 0
we have PAx P 1 0 2 0 101 0 0 1 Consider Lc 110 x 11 0 2 0 0 0 1 1 1
c 1 1 ; We have c 0 2
1 2 so 1 1
xc 0 0 101
1 1 1 100 Ux 1 11 110 LUx 1 1 1
100 1 1
so x 1
1 5...
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.
 Spring '08
 STAFF
 Math, Linear Algebra, Algebra

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