Linear Algebra Solutions 18

Linear Algebra Solutions 18 - 12. Also 1 0 A = E3 (2)E14...

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12. A = E 3 (2) E 14 E 42 (3) = E 3 (2) E 14 1 0 0 0 0 1 0 0 0 0 1 0 0 3 0 1 = E 3 (2) 0 3 0 1 0 1 0 0 0 0 1 0 1 0 0 0 = 0 3 0 1 0 1 0 0 0 0 2 0 1 0 0 0 . Also A - 1 = ( E 3 (2) E 14 E 42 (3)) - 1 = ( E 42 (3)) - 1 E - 1 14 ( E 3 (2)) - 1 = E 42 ( - 3) E 14 E 3 (1 / 2) = E 42 ( - 3) E 14 1 0 0 0 0 1 0 0 0 0 1 / 2 0 0 0 0 1 = E 42 ( - 3) 0 0 0 1 0 1 0 0 0 0 1 / 2 0 1 0 0 0 = 0 0 0 1 0 1 0 0 0 0 1 / 2 0 1 - 3 0 0 . 13. (All matrices in this question are over Z 2 .) (a) 1 1 0 1 0 0 1 1 1 1 1 1 1 0 0 1 f f f f f f f f 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 f f f f f f f f 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 1 0 0 0 0 1 0 0 0 1 1 f f f f f f f f 1 0 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 f f f f f f f f 0 0 0 1 1 0 0 1 1 0 1 0 1 1 1 0 21
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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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