Linear Algebra Solutions 15

Linear Algebra Solutions 15 - k 6 =-3, we see that B can be...

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Next suppose that a i = 0. Then diag( a 1 ,...,a n ) is row–equivalent to a matix containing a zero row and is hence singular. 3. [ A | I 3 ] = 0 0 2 1 2 6 3 7 9 f f f f f f 1 0 0 0 1 0 0 0 1 R 1 R 2 1 2 6 0 1 0 0 0 2 1 0 0 3 7 9 0 0 1 R 3 R 3 - 3 R 1 1 2 6 0 1 0 0 0 2 1 0 0 0 1 - 9 0 - 3 1 R 2 R 3 1 2 6 0 1 0 0 1 - 9 0 - 3 1 0 0 2 1 0 0 R 3 1 2 R 3 1 2 6 0 1 0 0 1 - 9 0 - 3 1 0 0 1 1 / 2 0 0 R 1 R 1 - 2 R 2 1 0 24 0 7 - 2 0 1 - 9 0 - 3 1 0 0 1 1 / 2 0 0 R 1 R 1 - 24 R 3 R 2 R 2 + 9 R 3 1 0 0 - 12 7 - 2 0 1 0 9 / 2 - 3 1 0 0 1 1 / 2 0 0 . Hence A is non–singular and A - 1 = - 12 7 - 2 9 / 2 - 3 1 1 / 2 0 0 . Also E 23 (9) E 13 ( - 24) E 12 ( - 2) E 3 (1 / 2) E 23 E 31 ( - 3) E 12 A = I 3 . Hence A - 1 = E 23 (9) E 13 ( - 24) E 12 ( - 2) E 3 (1 / 2) E 23 E 31 ( - 3) E 12 , so A = E 12 E 31 (3) E 23 E 3 (2) E 12 (2) E 13 (24) E 23 ( - 9) . 4. A = 1 2 k 3 - 1 1 5 3 - 5 1 2 k 0 - 7 1 - 3 k 0 - 7 - 5 - 5 k 1 2 k 0 - 7 1 - 3 k 0 0 - 6 - 2 k = B. Hence if - 6 - 2 k 6 = 0, i.e. if
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Unformatted text preview: k 6 =-3, we see that B can be reduced to I 3 and hence A is nonsingular. If k =-3, then B = 1 2-3-7 10 = B and consequently A is singu-lar, as it is rowequivalent to a matrix containing a zero row. 18...
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