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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #2 SOLUTIONS #1. Page 19; Exercise 4. If a + b c + d c d a b = 4 6 10 2 find a , b , c , and d . Solution: We equate the four entries of this 2 2 matrix: a + b = 4 c d = 10 c + d = 6 a b = 2 We can solve for a and c if we add the equations together: a + b = 4 a b = 2 2 a = 6 c + d = 6 c d = 10 2 c = 16 so that a = 3 and c = 8. From the equation a + b = 4 we find that b =, and from the equation c + d = 6 we find that d = 2. Hence a = 3, b = 1, c = 8, and d = 2 . #2. Page 20; Exercise 9. In Exercises 6 through 9, let A = 1 2 3 2 1 4 B = 1 0 2 1 3 2 C = 3 1 3 4 1 5 2 1 3 D = 3 2 2 4 E = 2 4 5 1 4 3 2 1 F = 4 5 2 3 If possible, compute the following: (a) (2 A ) T (b) ( A B ) T (c) ( 3 B T 2 A ) T (d) ( 3 A T 5 B T ) T (e) ( A ) T and ( A T ) (f) ( C + E + F T ) T Solution: (a) We must compute the transpose: 2 A = 2 4 6 4 2 8 = (2 A ) T = 2 4 4 2 6 8 1 2 MA 265 HOMEWORK ASSIGNMENT #2 SOLUTIONS (b) A is a 2 3 matrix, whereas B is a 3 2 matrix. Since they do not have the same size, it is impossible to compute A B , let alone ( A B ) T . (c) First we compute the transpose: B T = 1 2 3 0 1 2 = 3 B T = 3 6 9 0 3 6 . Hence we have the difference 3 B T 2 A = 3 6 9 0 3 6 2 4 6 4 2 8 = 1 2 3 4 1 2 = ( 3 B T 2 A ) T = 1 4 2 1 3 2 (b) A T is a 3 2 matrix, whereas B T is a 2 3 matrix. Since they do not have the same size, it is impossible to compute 3 A T 5 B T , let alone ( 3 A T 5 B T ) T . (e) We must compute the transpose A = 1 2 3 2 1 4 = ( A ) T =  1 2 2 1 3 4 Similarly, we compute A T = 1 2 2 1 3 4 =  ( A ) T =  1 2 2 1 3 4 (f) C and E are 3 3 matrices, whereas F T is a 2 2 matrix. Since they do not have the same size, it is impossible to compute C + E = F T , let alone ( C + E + F T ) T . #3. Page 31; Exercise 11. Consider the following matrices for Exercises 11 through 15: A = 1 2 3 2 1 4 B = 1 0 2 1 3 2 C = 3 1 3 4 1 5 2 1 3 D = 3 2 2 5 E = 2 4 5 1 4 3 2 1 F =  1 2 0 4 3 5 If possible, compute the following: (a) AB (b) B A (c) F T E (d) C B + D (e) AB + D 2 , where D 2 = D D . Solution: (a) We have the matrix product AB = 1 2 3 2 1 4 1 0 2 1 3 2 = 1(1) + 2(2) + 3(3) 1(0) + 2(1) + 3(2) 2(1) + 1(2) + 4(3) 2(0) + 1(1) + 4(2) = 14 8 16 9 MA 265 HOMEWORK ASSIGNMENT #2 SOLUTIONS 3 (b) We have the matrix product B A = 1 0 2 1 3 2 1 2 3 2 1 4 = 1(1) + 0(2) 1(2) + 0(1) 1(3) + 0(4) 2(1) + 1(2) 2(2) + 1(1) 2(3) + 1(4) 3(1) + 2(2) 3(2) + 2(1) 3(3) + 2(4) = 1 2 3 4 5 10 7 8 17 (c) We have the matrix product F T E = 1 0 3 2 4 5 2 4 5 1 4 3 2 1 = 1(2) + 0(0) + 3(3) 1( 4) + 0(1) + 3(2) 1(5) + 0(4) + 3(1) 2(2) + 4(0) + 5(3) 2( 4) + 4(1) + 5(2) 2(5) + 4(4) + 5(1) = 7 10 2 19 6 31 (d) First we compute the matrix product:...
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 Spring '10
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 Linear Algebra, Matrices, Vector Space

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