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Unformatted text preview: 18.06 Problem Set 6 Solutions Problem 1: Do problem 39 from section 5.3. Solution Recall that A 1 = C T /det ( A ). If we know det ( A ), then we get A 1 , hence find A . For the determinant, take determinants of both sides of the above equation. we have, det ( A 1 ) = 1 /det ( A ) = det ( C T ) /det ( A ) 4 , hence det ( A ) = det ( C ) 1 / 3 . We are done. Problem 2: Do problems 6 from section 6.1. Solution A is (lower) triangular, hence its eigenvalues are the entries on diagonal: 1 with multiplicity 2. Similarly B is (upper) triangular, hence its eigenvalue is 1 with multiplicity 2. The characteristic equations of AB and BA are both λ 2 4 λ + 1 = ( λ 2) 2 3 = 0, hence their eigenvalues are 2 ± √ 3. (a) The eigenvalues of AB are not the product of eigenvalues of A and B . (b) AB and BA have the same characteristic equation, hence the same eigen values. Problem 3: Problem 19 section 6.1....
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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.
 Fall '08
 Strang

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