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

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AMS-510 Solution for HW1 September 9 2013 Required Problems: 1.41, 1.45, 1,71 2.41, 2,55, 2,56, 2.69, 2.75, 2.79, 2.84, 2.85 3.53, 3.54 1.41 (a) 3 u - 2 v = 3(1 , - 2 , 4) - 2(3 , 5 , 1) = ( - 3 , - 16 , 10) (b) 5 u + 3 v - 4 w = 5(1 , - 2 , 4) + 3(3 , 5 , 1) - 4(2 , 1 , - 3) = (6 , 1 , 35) (c) u · v = (1 , - 2 , 4)(3 , 5 , 1) = 3 - 10 + 4 = - 3 u · w = (1 , - 2 , 4)(2 , 1 , - 3) = 2 - 2 - 12 = - 12 v · w = (3 , 5 , 1)(2 , 1 , - 3) = 6 + 5 - 3 = 8 1.45 (a) || u || = 1 + 4 + 16 = 21 || v || = 9 + 25 + 1 = 35 || w || = 4 + 1 + 9 = 14 1
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(b) || k u || = 9 , || k || · || u || = 3 || u || = 3 · 1 + 4 + 4 = 9 || u+v || = 120 , || u || + || v || = 3 + 13 = 16 1.71 (a) ( u + v ) w = (( u 1 ,...,u n ) + ( v 1 ,...,v n )) · ( w 1 ,...w n ) = ( u 1 + v 1 n + v n ) · ( w 1 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 ) = u · w + v · w (b) w ( u + v ) = ( w 1 n )(( u 1 n ) + ( v 1 n )) = ( w 1 n ) · ( u 1 + v 1 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 ) = w · u + w · v 2.41 (a) A T = ± 1 3 2 - 4 ² (b) B T = ± 5 - 6 0 7 ² (c) ( AB ) T = ± - 7 39 14 - 28 ² (d) A T B T = ± 5 15 10 - 40 ² 2
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2.55 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 2.56 AB = AC A - 1 AB = A - 1 AC B = C A = ± 1 2 2 2 ² B = ± 0 0 1 1 ² C = ± 2 2 0 0 ² AB = ± 2 2 2 2 ² AC = ± 2 2 2 2 ² 2.69 (a) ( A + A T ) T = A T + ( A T ) T = A T + A (b) ( A - A T ) T = A T
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AMS510HW1 - AMS-510 Solution for HW1 September 9 2013...

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