Linear Algebra Solutions 45

Linear Algebra Solutions 45 - A in the last equation gives...

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Consequently the complete solution of the system is x = 2 3 + 2 3 z, y = - 7 6 + 5 3 z , where z is arbitrary. Hence there are inFnitely many solutions. 10. Δ = f f f f f f f f 1 1 2 1 1 2 3 4 2 4 7 2 t + 6 2 2 6 - t t f f f f f f f f R 4 R 4 - 2 R 1 R 3 R 3 - 2 R 1 R 2 R 2 - R 1 = f f f f f f f f 1 1 2 1 0 1 1 3 0 2 3 2 t + 4 0 0 2 - t t - 2 f f f f f f f f = f f f f f f 1 1 3 2 3 2 t + 4 0 2 - t t - 2 f f f f f f R 2 R 2 - 2 R 1 = f f f f f f 1 1 3 0 1 2 t - 2 0 2 - t t - 2 f f f f f f = f f f f 1 2 t - 2 2 - t t - 2 f f f f = ( t - 2) f f f f 1 2 t - 2 - 1 1 f f f f = ( t - 2)(2 t - 1) . Hence Δ = 0 if and only if t = 2 or t = 1 2 . Consequently the given matrix B is non–singular if and only if t 6 = 2 and t 6 = 1 2 . 11. Let A be a 3 × 3 matrix with det A 6 = 0. Then (i) A adj A = (det A ) I 3 (1) (det A )det(adj A ) = det(det A · I 3 ) = (det A ) 3 . Hence, as det A 6 = 0, dividing out by det
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Unformatted text preview: A in the last equation gives det(adj A ) = (det A ) 2 . (ii) . Also from equation (1) µ 1 det A A ¶ adj A = I 3 , so adj A is non–singular and (adj A )-1 = 1 det A A. ±inally A-1 adj( A-1 ) = (det A-1 ) I 3 and multiplying both sides of the last equation by A gives adj( A-1 ) = A (det A-1 ) I 3 = 1 det A A. 48...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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