Previous answers holtlinalg2 33015 use an augmented

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4. 3/3 points | Previous Answers HoltLinAlg2 3.3.015. Use an augmented matrix and row operations to find A −1 , if it exists. (If the inverse does not exist, enter DNE into any cell.) [1, ­4, ­29, 62; 0, 1, 7, ­14; 0, 0, 1, ­1; 0, 0, 0, 1] Solution or Explanation so A = 1 4 1 5 0 1 7 7 0 0 1 1 0 0 0 1 , 1 4 1 5 1 0 0 0 0 1 7 7 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 R 4 + R 3 R 3 7 R 4 + R 2 R 2 5 R 4 + R 1 R 1 ~ 1 4 1 0 1 0 0 5 0 1 7 0 0 1 0 7 0 0 1 0 0 0 1 −1 0 0 0 1 0 0 0 1 7 R 3 + R 2 R 2 R 3 + R 1 R 1 ~ 1 4 0 0 1 0 −1 6 0 1 0 0 0 1 7 14 0 0 1 0 0 0 1 −1 0 0 0 1 0 0 0 1 4 R 2 + R 1 R 1 ~ 1 0 0 0 1 4 29 62 0 1 0 0 0 1 7 14 0 0 1 0 0 0 1 −1 0 0 0 1 0 0 0 1 = . 1 4 1 5 0 1 7 7 0 0 1 1 0 0 0 1 −1 1 4 29 62 0 1 7 14 0 0 1 −1 0 0 0 1
5. 3/3 points | Previous Answers HoltLinAlg2 3.3.024. Determine if the given linear transformation of T is invertible, and if so, find (HINT: Start by finding the matrix A such that If the inverse does not exist, enter DNE into all cells.)
6. 8/8 points | Previous Answers HoltLinAlg2 3.3.027. Let and be linear transformations given by Find the matrix A such that the following are true. (a) [7, 6; ­6, ­5] (b) [1, ­6; 0, 1] (c) [­5, ­6; 6, 7] (d) [1, 6; 0, 1] Solution or Explanation where and where (a) so Now so (b) so Now so (c) so Now so T 1 T 2 T 1 = T 2 = . x 1 x 2 2 x 1 + x 2 x 1 + x 2 x 1 x 2 8 x 1 + 7 x 2 x 1 + x 2 T 1 −1 ( T 2 ( x )) = A x A = T 1 ( T 2 −1 ( x )) = A x A = T 2 −1 ( T 1 ( x )) = A x A = T 2 ( T 1 −1 ( x )) = A x A = T 1 = T 1 ( x ) = A 1 x , x 1 x 2 A 1 = , 2 1 1 1 T 2 = T 2 ( x ) = A 2 x , x 1 x 2 A 2 = . 8 7 1 1 T 1 −1 ( T 2 ( x )) = A x = A 1 −1 A 2 x , A = A 1 −1 A 2 . A 1 −1 = = , 2 1 1 1 −1 1 −1 −1 2 A = = . 1 −1 −1 2 8 7 1 1 7 6 −6 −5 T 1 ( T 2 −1 ( x )) = A x = A 1 A 2 −1 x , A = A 1 A 2 −1 . A 2 −1 = = , 8 7 1 1 −1 1 7 −1 8 A = = . 2 1 1 1 1 7 −1 8 1 −6 0 1 T 2 −1 ( T 1 ( x )) = A x = A 2 −1 A 1 x , A = A 2 −1 A 1 . A 2 −1 = = , 8 7 1 1 −1 1 7 −1 8 A = = . 1 7 −1 8 2 1 1 1 −5 −6 6 7
(d) so Now so

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