assign5_soln

assign5_soln - Math 136 Assignment 5 Due: Wednesday, Feb...

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Math 136 Assignment 5 Due: Wednesday, Feb 24th 1. Let A = ± 12 - 1 3 - 2 - 1 ² , B = ± 1 0 0 02 - 3 ² , C = ± 5 - 1 - ² . Determine the following a) 2 A - B Solution: 2 A - B = ± 24 - 2 6 - 4 - 2 ² - ± 1 0 0 - 3 ² = ± 14 - 2 6 - 61 ² b) A ( B T + C T ) Solution: B T + C T = 10 0 - 3 + 51 - 1 - 1 22 = - 11 2 - 1 . So A ( B T + C T )= ± - 1 3 - 2 - 1 ² - 2 - 1 = ± 18 2 ² c) BA T + CA T Solution: T + T =( B + C ) A T =[ A ( B + C ) T ] T = ± 2 18 42 ² . 2. Prove that if ±x M (3 , 2) and a, b R are scalars, then ( a + b ) = a±x + b±x . Solution: Let = x 11 x 12 x 21 x 22 x 31 x 32 . Then ( a + b ) a + b ) x 11 x 12 x 21 x 22 x 31 x 32 = ( a + b ) x 11 ( a + b ) x 12 ( a + b ) x 21 ( a + b ) x 22 ( a + b ) x 31 ( a + b ) x 32 = ax 11 ax 12 ax 21 ax 22 ax 31 ax 32 + bx 11 bx 12 bx 21 bx 22 bx 31 bx 32 = a±x + b±x. 1
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2 3. Determine which of the following mappings are linear. Find the standard matrix of each linear mapping. a) f ( x 1 ,x 2 3 ) = ( x 1 + x 2 +1 3 , 0). Solution: Observe that f [(1 , 0 , 0)+(1 , 0 , 0)] = f (2 , 0 , 0) = (3 , 0 , 0), but f (1 , 0 , 0)+ f (1 , 0 , 0) = (2 , 0 , 0) + (2 , 0 , 0) = (4 , 0 , 0). Thus f is not linear. b) f ( x 1 2 ) = (0 1 +2 x 2 2 ). Solution: We have f [ c ( x 1 2 )+( y 1 ,y 2 )] = f ( cx 1 + y 1 , cx 2 + y 2 ) = (0 , ( cx 1 + y 1 ) + 2( cx 2 + y 2 ) , cx 2 + y 2 ) = c (0 1 x 2 2 ) + (0 1 y 2 2 ) = cf ( x 1 2 )+ f ( y 1 2 ) . Hence, f is linear. To ±nd the standard matrix we ±nd f (1 , 0) = (0 , 1 , 0) ,f (0 , 1) = (0 , 2 , 1) . Thus [ f ]= ± f (1 , 0) f (0 , 1) ² = 00 12 01 .
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This note was uploaded on 04/30/2010 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.

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assign5_soln - Math 136 Assignment 5 Due: Wednesday, Feb...

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