Winter 2006 - Nagy's Class - Final Exam

# Winter 2006 - Nagy's Class - Final Exam - Print Name...

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Print Name: Student Number: Section Time: Math 20F. Final Exam March 20, 2006 Read each question carefully, and answer each question completely. Show all of your work. No credit will be given for unsupported answers. Write your solutions clearly and legibly. No credit will be given for illegible solutions. 1. Consider the matrix A = 1 - 1 5 0 1 - 2 1 3 - 3 . (a) (5 Pts.) Find a basis for the subspace of all vectors b such that the linear system A x = b has solutions. Show your work. (b) (5 Pts.) Find a basis for the null space of A . Show your work. (c) (5 Pts.) Find a solution to the linear system A x = b with b = 1 1 5 . Is this solution unique? If yes, say why. If no, ±nd a second solution x with the same b . (a) 1 - 1 5 | b 1 0 1 - 2 | b 2 1 3 - 3 | b 3 1 - 1 5 | b 1 0 1 - 2 | b 2 0 4 - 8 | b 3 - b 1 1 - 1 5 | b 1 0 1 - 2 | b 2 0 0 0 | b 3 - b 1 - 4 b 2 The system is consistent if b 3 = b 1 + 4 b 2 , then all possible b have the form b = b 1 b 2 b 1 + 4 b 2 = 1 0 1 b 1 + 0 1 4 b 2 , therefore, the set W of all possible solution b is W = Span 1 0 1 , 0 1 4 . (b) 1 - 1 5 0 1 - 2 0 0 0 1 0 3 0 1 - 2 0 0 0 x 1 = - 3 x 3 x 2 = 2 x 3 x 3 is free. x = - 3 2 1 x 3 , N ( A ) = Span - 3 2 1 .

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(c) If b = 1 1 5 , then b 3 - b 1 - 4 b 2 = 5 - 1 - 4 = 0, hence the system is consistent. The null space of A is nontrivial, which says that there are inFnitely many solutions. 1 - 1 5 | 1 0 1 - 2 | 1 0 0 0 | 0 1 0 3 | 2 0 1 - 2 | 1 0 0 0 | 0 x 1 = 2 - 3 x 3 x 2 = 1 + 2 x 3 x 3 is free. The solutions are, x = 2 1 0 + - 3 2 1 x 3 , Two solutions can be obtained taking x 3 = 0 for one, and x 3 = 1 for the other solution.
2. Let u = · 1 1 ¸ , v = · 1 - 1 ¸ , and T : IR 2 IR 2 be a linear transformation given by T ( u ) = · 1 3 ¸ , T ( v ) = · 3 1 ¸ . (a) (5 Pts.) Find the matrix

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Winter 2006 - Nagy's Class - Final Exam - Print Name...

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