Linear Algebra Solutions 19

# Linear Algebra Solutions 19 - 1 0 → 0 0 0 1 0 0 0 0 1 0 0...

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Unformatted text preview: 1 0 → 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 . 0 0 1 0 0 1 1 0 0 1 Hence A is non–singular and A−1 1 0 (b) A = 1 1 1 1 0 1 0 1 1 0 14. (a) 1 1 = 1 1 1 0 1 1 1 1 . 0 0 1 11 0 1 1 R → R4 + R1 1 0 0 4 1 00 0 1 1 0 1 1 , so A is singular. 0 0 R3 → 1 R3 2 100 100 R1 → R1 − R3 0 1 0 010 R2 → R2 + R3 011 001 R1 ↔ R3 0 0 1/2 100 0 1 0 0 1 1/2 . 1 −1 −1 001 111 −1 1 0 200 R3 → R3 − R2 00 1/2 01 1/2 1 0 −1/2 Hence A−1 exists and A−1 (b) 224 1 0 1 010 R3 → R3 − 2R2 0 0 1/2 1 1/2 . = 0 1 −1 −1 100 0 1 0 001 101 0 1 0 002 101 R3 → 1 R3 0 1 0 2 001 R1 → R1 − 2R2 101 0 1 0 R1 ↔ R2 R2 ↔ R3 022 0 1 0 0 0 1 1 −2 −2 0 1 0 0 0 1 1/2 −1 −1 22 0 10 0 0 1 1 −2 0 ...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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