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HW12 - MA341 Homework 12 Solutions 9.3 1 1 1[A|I = 1 2 1 2...

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MA341 Homework 12 Solutions 9.3 11. [ A | I ] = 1 1 1 | 1 0 0 1 2 1 | 0 1 0 2 3 2 | 0 0 1 1 1 1 | 1 0 0 0 1 0 | - 1 1 0 0 1 0 | - 2 0 1 1 1 1 | 1 0 0 0 1 0 | - 1 1 0 0 0 0 | - 1 - 1 1 The matrix A is singular, hence the inverse of A doesn’t exist. 13. [ A | I ] = - 2 - 1 1 | 1 0 0 2 1 0 | 0 1 0 3 1 - 1 | 0 0 1 r 2 + r 1 r 3 + 3 2 r 1 - 2 - 1 1 | 1 0 0 0 0 1 | 1 1 0 0 - 1 2 1 2 | 3 2 0 1 r 3 × ( - 2) r 3 r 2 - 2 - 1 1 | 1 0 0 0 1 - 1 | - 3 0 - 2 0 0 1 | 1 1 0 r 1 - r 3 r 2 + r 3 - 2 - 1 0 | 0 - 1 0 0 1 0 | - 2 1 - 2 0 0 1 | 1 1 0 r 1 - r 2 - 2 0 0 | - 2 0 - 2 0 1 0 | - 2 1 - 2 0 0 1 | 1 1 0 r 1 / ( - 2) 1 0 0 | 1 0 1 0 1 0 | - 2 1 - 2 0 0 1 | 1 1 0 = [ I | A - 1 ] .
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MA341 Homework 12 Solutions 17. [ X ( t ) | I ] = e t e 4 t | 1 0 e t 4 e 4 t | 0 1 e t e 4 t | 1 0 0 3 e 4 t | - 1 1 e t 0 | 4 / 3 - 1 / 3 0 3 e 4 t | - 1 1 1 0 | 4 / 3 e - t - 1 / 3 e - t 0 1 | - 1 / 3 e - 4 t 1 / 3 e - 4 t = [ I | X - 1 ( t )] . 23. 1 0 0 3 1 2 1 5 - 2 = 1 1 2 5 - 2 = - 2 - 10 = - 12 25. 1 4 3 - 1 - 1 2 4 5 2 = 1 - 1 2 5 2 - 4 - 1 2 4 2 + 3 - 1 - 1 4 5 = 1( - 2 - 10) - 4( - 2 - 8) + 3( - 5 + 4) = - 12 + 40 - 3 = 25 31. x ( t ) = ( e 3 t ) (2 e 3 t ) ( - e 3 t ) = 3 e 3 t 6 e 3 t - 3 e 3 t 33. X ( t ) = ( e 5 t ) (3 e 2 t ) ( - 2 e 5 t ) ( - e 2 t ) = 5 e 5 t 6 e 2 t - 10 e 5 t - 2 e 2 t 35. x = 1 1 - 2 4 x , x ( t ) = e 3 t 2 e 3 t LHS = x ( t ) = ( e 3 t ) (2 e 3 t ) = 3 e 3 t 6 e 3 t ,
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MA341 Homework 12 Solutions and RHS = 1 1 - 2 4 x = 1 1 - 2 4 e 3 t 2 e 3 t = 3 e 3 t 6 e 3 t . Hence, x ( t ) provides a solution to the given system. 9.4 3. x ( t ) y ( t ) z ( t ) = t 2 - 1 1 0 0 e t t - 1 3 x ( t ) y ( t ) z ( t ) + t 5 - e t . 5. Let y = y 1 , y = y 2 , then the given equation can be written as y 2 ( t ) - 3 y 2 ( t ) - 10 y 1 ( t ) = sin t. Hence, the normal form is y 1 ( t ) = y 2 ( t ) , y 2 ( t ) = 3 y 2 ( t ) + 10 y 1 ( t ) + sin t, and the matrix form is y 1 ( t ) y 2 ( t ) = 0 1 10 3 y 1 ( t ) y 2 ( t ) + 0 sin t . 7. Let w = w 1 , w = w 2 , w = w 3 , and w = w 4 , then the given equation can be written as w 4 ( t ) + w 1 ( t ) = t 2 . Hence, the normal form is w 1 ( t ) = w 2 ( t ) , w 2 ( t ) = w 3 ( t ) , w 3 ( t ) = w 4 ( t ) , w 4 ( t ) = - w 1 ( t ) + t 2 , and the matrix form is w 1 ( t ) w 2 ( t ) w 3 ( t ) w 4 ( t ) = 0 1 0 0 0 0 1 0 0 0 0 1 - 1 0 0 0 w 1 ( t ) w 2 ( t ) w 3 ( t ) w 4 ( t ) + 0 0 0 t 2 .
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MA341 Homework 12 Solutions 13. Let x 1 ( t ) = t 3 and x 2 ( t ) = 4 1 , and assume that c 1 and c 2 are constants such that for every t in ( -∞ , ) we have: c 1 x 1 ( t ) + c 2 x 2 ( t ) = 0 c 1 t 3 + c 2 4 1 = 0 . Hence, we obtain the following linear homogeneous system with respect to c 1 and c 2 : tc 1 + 4 c 2 = 0 , 3 c 1 + c 2 = 0 . Its determinant is equal to t - 12. Therefore, there is a nontrivial solution for t = 12 and only trivial solution for t = 12. In other words, the vectors x 1 ( t ) = t 3 and x 2 ( t ) = 4 1 are linearly independent for t = 12 and linearly dependent for t = 12. 17. Let x 1 ( t ) = e 2 t 1 0 5 , x 2 ( t ) = e 2 t 1 1 - 1 , and x 3 ( t ) = e 3 t 0 1 0 . Assume that c 1 , c 2 , and c 3 are constants such that for every t in ( -∞ , ) we have: c 1 x 1 ( t ) + c 2 x 2 ( t ) + c 3 x 3 ( t ) = 0 c 1 e 2 t 1 0 5 + c 2 e 2 t 1 1 - 1 + c 3 e 3 t 0 1 0 = 0 .
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