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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS #1. Page 169; Exercise 2. Let A = 2 1 3 1 2 0 3 2 1 . (a) Find adj A . (b) Compute det( A ). (c) Verify Theorem 3.12; that is, show that A (adj A ) = (adj A ) A = det( A ) I 3 . Solution: (a) First we compute the cofactors of A . We have A 11 = ( 1) 1+1 2 0 2 1 = 2 A 21 = ( 1) 2+1 1 3 2 1 = 7 A 31 = ( 1) 3+1 1 3 2 0 = 6 A 12 = ( 1) 1+2 1 0 3 1 = 1 A 22 = ( 1) 2+2 2 3 3 1 = 7 A 32 = ( 1) 3+2 2 3 1 0 = 3 A 13 = ( 1) 1+3 1 2 3 2 = 4 A 23 = ( 1) 2+3 2 1 3 2 = 7 A 33 = ( 1) 3+3 2 1 1 2 = 5 Then the adjoint of A is the matrix adj A = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 = 2 7 6 1 7 3 4 7 5 (b) We can compute the determinant by expanding along the first row of A : det( A ) = a 11 A 11 + a 12 A 12 + a 13 A 13 = (2)(2) + (1)(1) + (3)( 4) = 7 (c) We have the matrix products A (adj A ) = 2 1 3 1 2 0 3 2 1 2 7 6 1 7 3 4 7 5 =  7 7 7 = det( A ) I 3 . (adj A ) A = 2 7 6 1 7 3 4 7 5 2 1 3 1 2 0 3 2 1 =  7 7 7 = det( A ) I 3 . #2. Page 169; Exercise 4. Find the inverse of the matrix in Exercise 2 by the method given in Corollary 3.4. 1 2 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS Solution: The inverse of the matrix is A 1 = 1 det( A ) (adj A ) =  2 7 1 6 7 1 7 1 3 7 4 7 1 5 7 #3. Page 172; Exercise 1. If possible, solve the following linear systems by Cramer’s rule: 2 x 1 + 4 x 2 + 6 x 3 = 2 x 1 + 2 x 3 = 2 x 1 + 3 x 2 x 3 = 5 Solution: Denote the coefficient matrix of the system by A = 2 4 6 1 0 2 2 3 1 = ⇒  A  = (2) 2 3 1 (4) 1 2 2 1 + (6) 1 0 2 3 = 26 . Then we have the values x 1 = 2 4 6 0 0 2 5 3 1  A  = (2) 2 3 1 (4) 2 5 1 + (6) 0 0 5 3 26 = 52 26 = 2 x 2 = 2 2 6 1 2 2 5 1  A  = (2) 2 5 1 (2) 1 2 2 1 + (6) 1 2 5 26 = 26 = 0 x 3 = 2 4 2 1 0 2 3 5  A  = (2) 3 5 (4) 1 2 5 + (2) 1 0 2 3 26 = 26 26 = 1 Hence the solution is x 1 = 2, x 2 = 0, and x 3 = 1. #4. Page 172; Exercise 3. Solve the following linear system for x 3 , by Cramer’s rule: 2 x 1 + x 2 + x 3 = 6 3 x 1 + 2 x 2 2 x 3 = 2 x 1 + x 2 + 2 x 3 = 4 Solution: Denote the coefficient matrix of the system by A = 2 1 1 3 2 2 1 1 2 = ⇒  A  = (2) 2 2 1 2 (1) 3 2 1 2 + (1) 3 2 1 1 = 5 . MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS 3 Then we have the value x 3 = 2 1 6 3 2 2 1 1 4  A  = (2) 2 2 1 4 (1) 3 2 1 4 + (6) 3 2 1 1 5 = 4 5 #5. Page 187; Exercise 2. Determine the head of the vector 2 5 whose tail is ( 3 , 2). Make a sketch. Solution: Say that the tail is P ( 3 , 2) and the head is Q ( x,y ). Then we have the directed line segment→ PQ = x ( 3) y 2 = 2 5 ....
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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.
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