AMS510HW1 - AMS-510 Solution for HW1 September 9 2013...

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3 - 2 = 3(1 , - 2 , 4) - 2(3 , 5 , 1) = ( - 3 , - 16 , 10) 5 + 3 - 4 = 5(1 , - 2 , 4) + 3(3 , 5 , 1) - 4(2 , 1 , - 3) = (6 , 1 , 35) · = (1 , - 2 , 4)(3 , 5 , 1) = 3 - 10 + 4 = - 3 · = (1 , - 2 , 4)(2 , 1 , - 3) = 2 - 2 - 12 = - 12 · = (3 , 5 , 1)(2 , 1 , - 3) = 6 + 5 - 3 = 8 || || = 1 + 4 + 16 = 21 || || = 9 + 25 + 1 = 35 || || = 4 + 1 + 9 = 14
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|| k || = 9 || k || · || || = 3 || || = 3 · 1 + 4 + 4 = 9 || || = 120 || || + || || = 3 + 13 = 16 ( + ) = (( u 1 , ..., u n ) + ( v 1 , ..., v n )) · ( w 1 , ...w n ) = ( u 1 + v 1 , ..., u n + v n ) · ( w 1 , ...w n ) = ( u 1 + v 1 ) + ... + ( u n + v n ) · w n = ( u 1 w 1 + ... + u n w n ) + ( v 1 w 1 + ... + v n w n ) = · + · ( + ) = ( w 1 , ...w n )(( u 1 , ..., u n ) + ( v 1 , ..., v n )) = ( w 1 , ...w n ) · ( u 1 + v 1 , ..., u n + v n ) = w n · ( u 1 + v 1 ) + ... + ( u n + v n ) = ( w 1 u 1 + ... + w n u n ) + ( w 1 v 1 + ... + w n v n ) = · + · A T = 1 3 2 - 4 B T = 5 - 6 0 7 ( AB ) T = - 7 39 14 - 28 A T B T = 5 15 10 - 40
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A - 1 = 1 1 2 1 2 5 1 3 7 - 1 = 1 1 1 2 - 5 3 - 1 2 - 1 B - 1 = 1 - 1 1 0 1 - 1 1 3 2 - 1 = 1 1 0 - 1 - 3 1 - 1 - 4 1 AB = AC A - 1 AB = A - 1 AC B = C A = 1 2 2 2 B = 0 0 1 1 C = 2 2 0 0 AB = 2 2 2 2 AC = 2 2 2 2 ( A + A T ) T = A T + ( A T ) T = A T + A ( A - A T ) T = A T - ( A T ) T = A t - A = - ( A - A T ) A = 1 2 ( A + A T ) + 1 2 ( A - A T ) B = 1 2 ( A + A T ) , C = 1 2 ( A - A T ) AA T = 3 - 4 4 3 3 4 - 4 3 = 25 0 0 25 = A T A BB T = 1 - 2 2 3 1 2 - 2 3 = 5 - 4 - 4 13 B T B = 1 2 - 2 3 1 - 2 2 3 = 5 4 4 13 CC T = 1 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 =
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