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HM8-soln

# HM8-soln - Solution for Homework 8 December 5 2011 3.65(a 1...

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Solution for Homework 8 December 5, 2011 3.65 (a) E 1 = 1 0 0 0 0 1 0 1 0 E 2 = 1 0 0 0 3 0 0 0 1 E 3 = 1 0 2 0 1 0 0 0 1 (b) e - 1 1 : Interchange R 2 and R 3 ; e - 1 2 : Replace R 2 by 1 3 R 2 ; e - 1 3 : Replace R 1 by - 2 R 3 + R 1 . E 0 1 = 1 0 0 0 0 1 0 1 0 = E - 1 1 E 0 2 = 1 0 0 0 1 3 0 0 0 1 = E - 1 2 E 0 3 = 1 0 - 2 0 1 0 0 0 1 = E - 1 3 . (c) f 1 : Interchange C 2 and C 3 ; f 2 : Replace C 2 by 3 C 2 ; f 3 : Replace C 1 by 2 C 3 + C 1 . (d) F 1 = 1 0 0 0 0 1 0 1 0 = E T 1 F 2 = 1 0 0 0 3 0 0 0 1 = E T 2 F 3 = 1 0 0 0 1 0 2 0 1 = E T 3 . 3.66 A 1 2 3 4 r 2 - 3 r 1 1 2 0 - 2 r 1 + r 2 1 0 0 - 2 Therefore 1 1 0 1 1 0 - 3 1 A = 1 0 0 - 2 or A = 1 0 - 3 1 - 1 1 1 0 1 - 1 1 0 0 - 2 = 1 0 3 1 1 - 1 0 1 1 0 0 - 2 1

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B 3 - 6 - 2 4 r 2 + 2 3 r 1 3 - 6 0 0 r 1 × 1 3 1 - 2 0 0 c 2 +2 c 1 1 0 0 0 Therefore, 1 3 0 0 1 1 0 2 3 1 B 1 2 0 1 = 1 0 0 0 or B = 1 0 2 3 1 - 1 1 3 0 0 1 - 1 1 0 0 0 1 2 0 1 - 1 = 1 0 - 2 3 1 3 0 0 1 1 0 0 0 1 - 2 0 1 C 2 6 - 3 - 7 r 2 + 3 2 r 1 2 6 0 2 r 1 - 3 r 2 2 0 0 2 r 1 × 1 2 1 0 0 2 Therefore, 1 2 0 0 1 1 - 3 0 1 1 0 3 2 1 C = 1 0 0 2 or C = 1 0 3 2 1 - 1 1 - 3 0 1 - 1 1 2 0 0 1 - 1 1 0 0 2 = 1 0 - 3 2 1 1 3 0 1 2 0 0 1 1 0 0 2 D 1 2 0 0 1 3 3 8 7 r 3 - 3 r 1 1 2 0 0 1 3 0 2 7 r 3 - 2 r 2 1 2 0 0 1 3 0 0 1 r 2 - 3 r 3 1 2 0 0 1 0 0 0 1 Therefore, 1 0 0 0 1 - 3 0 0 1 1 0 0 0 1 0 0 - 2 1 1 0 0 0 1 0 - 3 0 1 D = 1 2 0 0 1 0 0 0 1 or D = 1 0 0 0 1 0 - 3 0 1 - 1 1 0 0 0 1 0 0 - 2 1 - 1 1 0 0 0 1 - 3 0 0 1 - 1 1 2 0 0 1 0 0 0 1 = 1 0 0 0 1 0 3 0 1 1 0 0 0 1 0 0 2 1 1 0 0 0 1 3 0 0 1 1 2 0 0 1 0 0 0 1 2
3.67 A 1 - 2 - 1 1 0 0 2 - 3 1 0 1 0 3 - 4 4 0 0 1 r 2 - 2 r 1 1 - 2 - 1 1 0 0 0 1 3 - 2 1 0 3 - 4 4 0 0 1 r 3 - 3 r 1 1 - 2 - 1 1 0 0 0 1 3 - 2 1 0 0 2 7 - 3 0 1 r 3 - 2 r 2 1 - 2 - 1 1 0 0 0 1 3 - 2 1 0 0 0 1 1 - 2 1 r 2 - 3 r 3 1 - 2 - 1 1 0 0 0 1 0 - 5 7 - 3 0 0 1 1 - 2 1 r 1 + r 3 1 - 2 0 2 - 2 1 0 1 0 - 5 7 - 3 0 0 1 1 - 2 1 r 1 +2 r 2 1 0 0 - 8 12 - 5 0 1 0 - 5 7 - 3 0 0 1 1 - 2 1 A - 1 = - 8 12 - 5 - 5 7 - 3 1 - 2 1 . Using the same technique we can see, B has no inverse, C - 1 = 29 2 - 17 2 7 2 - 5 2 3 2 - 1 2 3 - 2 1 , D - 1 = 8 - 3 - 1 - 5 2 1 10 - 4 - 1 3.68 A - 1 = 1 - 1 1 - 1 · · · ( - 1) n - 1 1 - 1 1 - 1 . . . . . . . . . . . . . . . 1 - 1 1 - 1 1 - 1 1 1 - 1 1 B - 1 = 1 - 1 1 - 1 1 - 1 . . . . . . 1 - 1 1 - 1 1 4.71 (a) E 1 = 4(5 u - 6 v ) + 2(3 u + v ) = 26 u - 22 v 3

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(b) 7 v + 8 is not defined, therefore we cannot get E 2 . (c) E 3 = 6(3 u + 2 v ) + 5 u - 7 v = 23 u + 5 v (d) 2 /v is not defined, so we cannot get E 4 . 4.74 (a) ( a, b ) + ( c, d ) = ( a + d, b + c ) 6 = ( c + b, d + a ) = ( c, d ) + ( a, b ) (b) m ( a, b ) + n ( a, b ) = ( a, b ) + ( a, b ) = (2 a, 2 b ) 6 = ( a, b ) = ( m + n )( a, b ) (c) m ( a, b )+ n ( a, b ) = ( ma, mb )+( na, nb ) = (0 , 0) 6 = (( m + n ) a, ( m + n ) b ) = ( m + n )( a, b ) (d) m ( a, b )+ n ( a, b ) = ( ma, mb )+( na, nb ) = ( mna 2 , mnb 2 ) 6 = (( m + n ) a, ( m + n ) b ) = ( m + n )( a, b ) 4.75 Proof:A1 [( a 1 , a 2 , · · · ) + ( b 1 , b 2 , · · · )] + ( c 1 , c 2 , · · · ) = ( a 1 + b 1 , a 2 + b 2 , · · · ) + ( c 1 , c 2 , · · · ) = ( a 1 + b 1 + c 1 , a 2 + b 2 + c 2 , · · · ) = ( a 1 , a 2 , · · · ) + ( b 1 + c 1 , b 2 + c 2 ,
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HM8-soln - Solution for Homework 8 December 5 2011 3.65(a 1...

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