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homework set 8-Solutions - BILKENT UNIVERSITY Department of...

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B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 8 U. Mu˘gan July 2, 2008 Homework problems from the 2 nd Edition, SECTION 3.4 5 (5) ) 2 2 - 1 3 2 ‚ • - 4 2 1 3 = - 9 1 - 10 12 , - 4 2 1 3 ‚ • 2 - 1 3 2 = - 2 8 11 5 . 8 (8) ) 1 0 3 2 - 5 4 3 0 - 1 4 6 5 = 21 15 34 0 , 3 0 - 1 4 6 5 1 0 3 2 - 5 4 = 3 0 9 7 - 20 13 16 - 25 38 . 11 (11) ) AB = £ 3 - 5 / 2 7 5 6 - 1 4 2 3 = £ 11 1 5 3 / . But, the product BA is not defined. 15 (15) ) A ( BC ) = 3 2 £ 1 - 1 2 / 2 0 0 3 1 4 = 3 2 £ 4 5 / = 12 15 8 10 . ( AB ) C = 3 2 £ 1 - 1 2 / 2 0 0 3 1 4 = 3 - 3 6 2 - 2 4 2 0 0 3 1 4 = 12 15 8 10 . 17 (17) ) Since the coefficient matrix is already is in echelon form, so we can write down the solutions by the back substitution. x 3 = s, x 4 = t, x 2 = - 2 s + 7 t, x 1 = 5 s - 4 t. In vector form ~x = s (5 , - 2 , 1 , 0) + t ( - 4 , 7 , 0 , 1) .
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