# –/2 pointsHoltLinAlg1 51004 Find for the given matrix

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Current Score : – / 41 Due : Friday, November 21 2014 10:00 PM PST 1. –/2 pointsHoltLinAlg1 5.1.004. Find for the given matrix A. M 23 = [0, -6, 0; 0, 6, 0; 4, 2, 2] M 31 = [-6, 1, 0; 3, -3, 0; 2, 3, 2] Solution or Explanation UW Common Math 308 Section 5.1 (Homework) YIYAO LIU Math 308, section E, Fall 2014 Instructor: Richard Balka WebAssign The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension M 23 and M 31 A = 0 6 1 0 5 3 3 0 0 6 3 0 4 2 3 2 M 23 = M 31 = 0 6 0 0 6 0 4 2 2 6 1 0 3 3 0 2 3 2
2. –/2 pointsHoltLinAlg1 5.1.008. Find C 13 and C 22 for the given matrix A . C 13 = (No Response) 58 C 22 = (No Response) 27 Solution or Explanation 9 3 0 5 3 6 6 8 3 C 13 = ( 1) 1+3 | M 13 | = = ( 5 )( 8 ) ( 3 )( 6 ) = 58 C 22 = ( 1) 2+2 | M 22 | = = ( 9 )( 3 ) (0)( 6 ) = 27 5 3 6 8 9 0 6 3
3. –/4 pointsHoltLinAlg1 5.1.014. Find the determinant for the given matrix A in two ways, by using cofactor expansion along the indicated row or column. (a) along the first row det( A ) = A = 5 1 3 0 1 5 0 1 4 5 0 1 0 1 5 0
(b) along the third column det( A ) = (No Response) -6 Use the determinant to decide if is invertible. Since A (No Response) is invertible, and hence T (No Response) is invertible. Solution or Explanation (a) Using the first row, (b) Using the third column, Since A is invertible, and hence T is invertible. T ( x ) = A ( x ) det( A ) (No Response) 0, det( A ) = a 11 C 11 + a 12 C 12 + a 13 C 13 + a 14 C 14 = ( 5 ) C 11 + (1) C 12 + ( 3 ) C 13 + (0) C 14 = ( 5 )( 1) 1 + 1 | M 11 | + (1)( 1) 1 + 2 | M 12 | + ( 3 )( 1) 1 + 3 | M 13 | = ( 5 ) (1) + ( 3 ) = 6 . 5 0 1 5 0 1 1 5 0 1 0 1 4 0 1 0 5 0 1 5 1 4 5 1 0 1 0 det( A ) = a 13 C 13 + a 23 C 23 + a 33 C 33 + a 43 C 43 = ( 3 ) C 13 + (0) C 23 + (0) C 33 + ( 5 ) C 43 = ( 3 )( 1) 1 + 3 | M 13 | + ( 5 )( 1) 4 + 3 | M 43 | = ( 3 ) ( 5 ) = 6 . 1 5 1 4 5 1 0 1 0 5 1 0 1 5 1 4 5 1 det( A ) 0,
4. –/2 pointsHoltLinAlg1 5.1.030. Find all real values of a such that the given matrix is not invertible. (HINT: Think determinants, not row operations. Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Solution or Explanation has no solutions A is invertible for all 5. –/3 pointsHoltLinAlg1 5.1.034.