{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework_6_solutions

# homework_6_solutions - MA 265 HOMEWORK ASSIGNMENT#6...

This preview shows pages 1–5. Sign up to view the full content.

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 0 0 0 - 7 0 0 0 - 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 0 0 0 - 7 0 0 0 - 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = 0 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) 0 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) 0 2 3 - 1 - (4) 0 2 - 5 - 1 + (6) 0 0 - 5 3 26 = - 52 26 = - 2 x 2 = 2 2 6 1 0 2 2 - 5 - 1 | A | = (2) 0 2 - 5 - 1 - (2) 1 2 2 - 1 + (6) 1 0 2 - 5 26 = 0 26 = 0 x 3 = 2 4 2 1 0 0 2 3 - 5 | A | = (2) 0 0 3 - 5 - (4) 1 0 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 . Since x + 3 = - 2 and y - 2 = 5, we see that x = - 5 and y = 7. Hence the head is ( - 5 , 7). A sketch can be found below. y ( - 5 , 7) - 7 - - 5 - - 3 U , , , , , , , , , , , , , , , , , , , , , , , , ( - 3 , 2) - - 1 / | - 5 | | - 3 | | - 1 | 1 | | 3 | | 5 x O #6. Page 187; Exercise 5. For what values of a and b are the vectors a - b 2 and 4 a + b equal? Solution: Upon equating both components, we find the system of equations a - b = 4 a + b = 2 Adding these two equations together gives 2 a = 6, so that a = 3. Similarly, subtracting the two equations gives - 2 b = 2, so that b = - 1. Hence the values are a = 3 and b = - 1.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS #7. Page 187; Exercise 7. In Exercises 7 and 8, determine the components of each vector --→ PQ .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}