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**Unformatted text preview: **MATH 51 MIDTERM 2 SOLUTIONS November 13, 2008 1(a) . Find the matrix for for the linear map T given by T x y = 7 x- 2 y x + 3 y 5 y . Solution : By inspection, matrix is 7- 2 1 3 5 . Alternatively, one can find the matrix by calculating T ( e 1 ) and T ( e 2 ) to get columns 1 and 2, respectively. 1(b) . Find the matrix for reflection in R 2 about the line y =- x . Solution : First column is T ( e 1 ) =- e 2 =- 1 . Second column is T ( e 2 ) =- e 1 =- 1 . Thus the matrix is- 1- 1 . 1(c) . Find the matrix for T : R 3 R 3 , where T is rotation by 180 about the y-axis, followed by rotation by 180 about the z-axis. Solution : e 1 - e 1 e 1 e 2 e 2 - e 2 e 3 - e 3 - e 3 , so the columns of the matrix are e 1 ,- e 2 , and- e 3 . Thus the matrix is 1- 1- 1 2 . Find the determinant of the matrix C = 1 2 3- 1- 1 1 2 1 1 . Solution : 1 2 3- 1- 1 1 2 1 1 = 1 2 3 1 4- 3- 5 = 1 4- 3- 5 = 1(- 5)- 4(- 3) = 7, where we added row 1 to row 2 and subtracted 2 times row 1 from row 3, and then expanded using the first column. Instead of expanding using the first column we could also add 3 times row 2 to row 3: 1 2 3 1 4- 3- 5 = 1 2 3 0 1 4 0 0 7 = 1 1 7 = 7 . 3. Find the inverse of the matrix A = 0 0 1 0 1 1 1 1 1 . 1 Solution : 0 0 1 | 1 0 0 0 1 1 | 0 1 0 1 1 1 | 0 0 1 1 1 1 | 0 0 1 0 1 1 | 0 1 0 0 0 1 | 1 0 0 1 1 0 | - 1 0 1 0 1 0 | - 1 1 0 0 0 1 | 1 0 0 1 0 0 |- 1 1 0 1 0 | - 1 1 0 0 1 | 1 so A- 1 is - 1 1- 1 1 1 ....

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