assign5_soln

# assign5_soln - Math 136 Assignment 5 Due Wednesday Feb 24th...

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Unformatted text preview: Math 136 Assignment 5 Due: Wednesday, Feb 24th 1. Let A = 1 2- 1 3- 2- 1 , B = 1 0 0 2- 3 , C = 5- 1 2 1- 1 2 . Determine the following a) 2 A- B Solution: 2 A- B = 2 4- 2 6- 4- 2- 1 0 0 2- 3 = 1 4- 2 6- 6 1 b) A ( B T + C T ) Solution: B T + C T = 1 2- 3 + 5 1- 1- 1 2 2 = 6 1- 1 1 2- 1 . So A ( B T + C T ) = 1 2- 1 3- 2- 1 6 1- 1 1 2- 1 = 2 4 18 2 c) BA T + CA T Solution: BA T + CA T = ( B + C ) A T = [ A ( B + C ) T ] T = 2 18 4 2 . 2. Prove that if ~x ∈ M (3 , 2) and a, b ∈ R are scalars, then ( a + b ) ~x = a~x + b~x . Solution: Let ~x = x 11 x 12 x 21 x 22 x 31 x 32 . Then ( a + b ) ~x = ( 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 2 3. Determine which of the following mappings are linear. Find the standard matrix of each linear mapping. a) f ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 + 1 , x 3 , 0). Solution: Observe that f [(1 , , 0)+(1 , , 0)] = f (2 , , 0) = (3 , , 0), but f (1 , , 0)+ f (1 , , 0) = (2 , , 0) + (2 , , 0) = (4 , , 0). Thus f is not linear. b) f ( x 1 , x 2 ) = (0 , x 1 + 2 x 2 , x 2 ). Solution: We have f [ c ( x 1 , x 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 , x 1 + 2 x 2 , x...
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## This note was uploaded on 05/04/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.

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assign5_soln - Math 136 Assignment 5 Due Wednesday Feb 24th...

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