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Unformatted text preview: 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 3 2 5 4 ‚ 3 1 4 6 5 = • 21 15 34 ‚ , 3 1 4 6 5 • 1 3 2 5 4 ‚ = 3 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.by the back substitution....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

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