This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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:...
View
Full
Document
This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
 Spring '10
 ...

Click to edit the document details