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math2j_pracmid_solution

math2j_pracmid_solution - Math 2J Practice Midterm Solution...

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Math 2J Practice Midterm Solution 1. Solve the linear system 2 3 1 2 3 1 2 3 3 3 5 9 2 2 3 3 x x x x x x x x + = + + = - + + = Solution : 2. Find the inverse of matrix 1 4 3 2 7 6 1 7 2 A - = - - - by the following methods (a) Augmented with 3 I and use reduced row echelon form to obtain 1 A - Solution : (b) Use the formula involving det( ) and A adjA . Solution : ( 29 ( 29 ( 29 3 6 7 3 4 1 13 2 7 3 4 1 28 2 7 6 7 1 1 3 31 1 2 21 1 1 11 = - - - = - = - - - = - = - - - = + + + A A A - 3 2 3 3 2 1 9 5 3 1 1 0 3 1 R R - 3 2 3 1 1 0 9 5 3 3 2 1 2 1 3 R R + - - - 3 11 3 1 1 0 0 1 0 3 2 1 3 2 R R + - - - 8 11 3 1 0 0 0 1 0 3 2 1 3 3 2 11 8 3 2 1 2 3 = + + = - = x x x x x = 3 11 5 3 2 1 x x x - - - - 1 0 0 0 1 0 0 0 1 2 7 1 6 7 2 3 4 1 + - + 3 1 2 1 2 R R R R - - 1 0 1 0 1 2 0 0 1 1 3 0 0 1 0 3 4 1 + - 3 2 3 R R - - - 1 3 7 0 1 2 0 0 1 1 0 0 0 1 0 3 4 1 + 1 3 3 R R - - - - 1 3 7 0 1 2 3 9 20 1 0 0 0 1 0 0 4 1 + - 1 2 4 R R - - - - = - - - - - 1 3 7 0 1 2 3 13 28 1 3 7 0 1 2 3 13 28 1 0 0 0 1 0 0 0 1 1 A
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( 29 ( 29 ( 29 0 6 2 3 1 1 1 2 1 3 1 1 2 2 1 6 2 1 2 3 32 2 2 22 2 1 12 = - - - = = - - - = = - - - = + + + A A A ( 29 ( 29 ( 29 1 7 2 4 1 1 3 7 1 4 1 1 7 7 1 7 2 1 3 3 33 3 2 23 3 1 13 = - - - = - = - = - = - - - = + + + A A A - - - - = 1 3 7 0 1 2 3 13 28 adjA , and 1 21 8 28 7 1 7 2 ) 3 ( 2 1 6 2 4 2 7 6 7 1 ) det( = + + - = - - - + - - - - - = A - - - - = = - 1 3 7 0 1 2 3 13 28 ) det( 1 1 adjA A A 3. Show that the block matrix 0 I A I is invertible and find its inverse. Solution : Note: for a block matrix = D C B A M , ) det( ) det( BC AD M - = . ( 29 M I I A I M = = = 0 1 det 0 ) det( 2 is invertible. If M has an inverse the it must have the form = - 3 3 2 1 1 M M M M M We know that = + + = = - I I M AM M AM M M M M M M I A I MM 0 0 0 4 2 3 1 2 1 4 3 2 1 1 - = = = + - = = + = + = = - I A I M I M I M AM A M M A M AM M I M 0 0 0 0 1 4 4 2 3 3 3 1 2 1 4. Combine the methods of row reduction and cofactor expansion to compute the determinant of matrix
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3 2 1 4 1 3 0 3 3 4 2 8 3 4 0 4 A - - - - = - - - Solution: 5. Use Cramer’s rule to compute the solutions of the system 1 2 3 1 3 1 2 3 2 4 2 2 3 3 3 x x x x x x x x + + = - + = + + = - Solution : { } 1 1 4 1 1 2 0 2 ( ) 1
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