HW12 - MA341 Homework 12 Solutions 9.3 11. [ A | I ] = 1 1...

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Unformatted text preview: 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 1 0 | - 1 1 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 1 0 3 1- 1 | 0 0 1 → r 2 + r 1 r 3 + 3 2 r 1 - 2- 1 1 | 1 0 0 1 | 1 1 0- 1 2 1 2 | 3 2 0 1 → r 3 × (- 2) r 3 ↔ r 2 - 2- 1 1 | 1 1- 1 | - 3 0- 2 1 | 1 1 → r 1- r 3 r 2 + r 3 - 2- 1 0 |- 1 1 | - 2 1- 2 1 | 1 1 → r 1- r 2 - 2 0 0 | - 2 0- 2 1 0 | - 2 1- 2 0 1 | 1 1 → r 1 / (- 2) 1 0 0 | 1 1 0 1 0 | - 2 1- 2 0 0 1 | 1 1 = [ I | A- 1 ] . 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 3 e 4 t | - 1 1 # → " e t | 4 / 3- 1 / 3 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 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 # , 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 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 ) # = " 1 10 3 #" y 1 ( t ) y 2 ( t ) # + " sin t # . 7. Let w = w 1 , w = w 2 , w 00 = w 3 , and w 000 = 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 ) = 1 0 0 0 1 0 0 0 1- 1 0 0 0 w 1 ( t ) w 2 ( t ) w 3 ( t ) w 4 ( t ) + t 2 . 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 (-∞ ,...
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This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.

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HW12 - MA341 Homework 12 Solutions 9.3 11. [ A | I ] = 1 1...

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