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# 2.5 - Math 2940 Solutions Fall 2011 Section 2.5 001 2 0 1 0...

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Math 2940 Solutions, Fall 2011 Section 2 . 5 2 ) 0 0 1 0 1 0 1 0 0 looks like the identity, except the first and third rows (or columns) are switched. So switching them again should be the identity. You can check 0 0 1 0 1 0 1 0 0 - 1 = 0 0 1 0 1 0 1 0 0 . P = 0 1 0 0 0 1 1 0 0 looks like the identity, except the rows (or columns) are permuted backwards (first to third, third to second, second to first). So doing this two more times should be the identity. You can check 0 1 0 0 0 1 1 0 0 - 1 = 0 1 0 0 0 1 1 0 0 2 = 0 0 1 1 0 0 0 1 0 . 3 ) We have 10 x + 20 y = 1 and 20 x + 50 y = 0. Adding - 2 times the first equation to the second gives 10 y = - 2 so y = - 1 / 5 and then x = 1 / 2 so the first column of A - 1 is 1 / 2 - 1 / 5 . Solving We have 10 t + 20 z = 0 and 20 t + 50 z = 1. Adding - 2 times the first equation to the second gives 10 z = 1 so z = 1 / 10 and then t = - 1 / 5 so the first column of A - 1 is - 1 / 5 1 / 10 . We have A - 1 = 1 / 2 - 1 / 5 - 1 / 5 1 / 10 . 5 ) - 1 0 1 0 1 0 0 0 1 2 = 1 0 0 0 1 0 0 0 1 .

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