24_hwc1SolnsODDA

# 24_hwc1SolnsODDA - A is not invertible SOLUTION Write A in...

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3.19 Given an example of noninvertible matrices A and B such that A + B is invertible. SOLUTION ± 1 0 0 1 ² = ± 1 0 0 0 ² + ± 0 0 0 1 ² . 3.21 Let A B and C be three invertible 2 × 2 matrices. Let D = ABC . Suppose that D - 1 = h 4 5 5 6 i , A = h 5 7 7 10 i and C - 1 = h 1 1 1 2 i . Compute B . SOLUTION Starting from D = ABC , and multiplying through by A - 1 on the left, and then C - 1 on the right, we have B = A - 1 DC - 1 . (The order is important!) Next, using the formula for the inverse of a 2 × 2 matrix from Theorem 7, A - 1 = ± 10 - 7 - 7 5 ² and D = ± - 6 5 5 - 4 ² . We can now multiply things out: A - 1 D = ± - 95 78 67 - 55 ² , and then we ﬁnd B = ( A - 1 D ) C - 1 = ± - 17 61 12 - 43 ² . 3.23 Let A be a 3 × 3 matrix. Show that if the ﬁrst column minus the second column equals the third column, then
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Unformatted text preview: A is not invertible. SOLUTION Write A in the form A = [ v 1 , v 2 , v 3 ]. Then the given information is that v 1-v 2 = v 3 . This is the same as v 1-v 2-v 3 = 0, and this is the same as [ v 1 , v 2 , v 3 ] 1-1-1 = 0. Hence there is a non zero vector v with A v = 0, and this means that A is not invertible. 3.25 (a) Show that h 1 1 /n 1 i n is the same matrix for all values of n . (b) Compute the limit lim n →∞ h 1 1 /n 3 1 i n 21 /september/ 2005; 22:09 25...
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## This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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