exam1soln

exam1soln - 18.06 Quiz 1 Solutions October 5, 2009 1 2 3...

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18.06 Quiz 1 Solutions October 5, 2009 Problem 1. Consider the matrix A = 1 2 3 1 1 1 0 1 3 . (a) Find the factorization A = LU . (b) Find the inverse of A . (c) For which values of c is the matrix 1 2 3 1 1 1 0 1 c invertible? Solution (a) We row reduce A by subtracting row 1 from row 2 ( E 12 ) and then add row 2 to row 3 ( E 23 ) to find the upper triangular matrix U = 1 2 3 0 - 1 - 2 0 0 1 . Since we can reverse this process and subtract row 2 from row 3 in U , followed by adding row 1 to row 2 to obtain A , we see that the lower triangular matrix is the product: 1 0 0 1 1 0 0 0 1 1 0 0 0 1 0 0 - 1 1 = 1 0 0 1 1 0 0 - 1 1 . Hence we find A = 1 0 0 1 1 0 0 - 1 1 1 2 3 0 - 1 - 2 0 0 1 .
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A - 1 = ( LU ) - 1 = U - 1 L - 1 . We explicitly compute U - 1 and find: 1 2 3 | 1 0 0 0 - 1 - 2 | 0 1 0 0 0 1 | 0 0 1 1 2 0 | 1 0 - 3 0 - 1 0 | 0 1 2 0 0 1 | 0 0 1 1 0 0 | 1 2 1 0 1 0 | 0 - 1 - 2 0 0 1 | 0 0 1 . Similary, we compute L - 1 = 1 0 0 - 1 1 0 - 1 1 1 . Therefore A - 1 = 1 2 1 0 - 1 - 2 0 0 1 1 0 0 - 1 1 0 - 1 1 1 = - 2 3 1 3 - 3 - 2 - 1 1 1 . (c) We row reduce to find : 1 2 3 1 1 1 0 1 c 1 2 3 0 - 1 - 2 0 0 c - 2 . Note that A is invertible if and only if it has 3 nonzero pivots. Thus A is invertible when c 6 = 2. Problem 2.
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Fall '08 term at MIT.

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exam1soln - 18.06 Quiz 1 Solutions October 5, 2009 1 2 3...

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