Linear Algebra Solutions 20

# Linear Algebra Solutions 20 - R1 → R1 − R3 100 0 1 0...

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Unformatted text preview: R1 → R1 − R3 100 0 1 0 001 Hence A−1 exists and −1/2 2 1 0 0 1 . 1/2 −1 −1 (c) −1/2 0 A−1 = 1/2 4 6 −3 4 1 R → 7 R2 0 0 7 2 0 R3 → 1 R3 5 00 5 0 Hence A is singular by virtue of 2 00 100 0 1 0 (d) 0 −5 0 0 07 001 2 1 0 1 . −1 −1 4 6 −3 6 −3 1 . 0 1 R3 → R3 − R2 0 0 00 0 0 1 the zero row. R1 → 1 R1 100 2 −1 0 1 0 R2 → 5 R2 R3 → 1 R3 001 7 Hence A−1 exists and A−1 = diag (1/2, −1/5, 1/7). 1/2 0 0 0 −1/5 0 . 0 0 1/7 (Of course this was also immediate from Question 2.) 1246 1000 1006 0 1 2 0 0 1 2 0 0 1 0 0 R → R1 − 2R2 (e) 0 0 1 2 0 0 1 2 0 0 1 0 1 0002 0001 0002 1 −2 00 100 6 0 1 0 −4 0 1 −2 0 R2 → R2 − 2R3 0 0 1 0 0 1 0 2 0 0 01 000 2 1 −2 0 −3 1000 R1 → R1 − 3R4 0 1 −2 2 R2 → R2 + 2R4 0 1 0 0 . 0 0 1 0 0 0 1 −1 R3 → R3 − R4 0 0 0 1/2 0001 R4 → 1 R4 2 Hence A−1 exists and A−1 1 −2 0 −3 0 1 −2 2 . = 0 0 1 −1 0 0 0 1/2 23 1 −2 0 0 0 1 0 0 0 0 1 0 0 001 ...
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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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