3e-spring2011-exam_1_solutions

3e-spring2011-exam_1_solutions - Math 3E — Exam 1...

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Unformatted text preview: Math 3E — Exam 1 Solutions Laney College, Spring 2011 Fred Bourgoin 1. (10 points) Use matrices to solve the system of linear equations. x 1 + 3 x 2 + 2 x 3 + x 4 = 2 2 x 1 + x 2 + x 3 − x 4 = 3 x 1 − 2 x 2 − x 3 − 2 x 4 = 1 Answer. Reduce the augmented matrix to its rref. 1 3 2 1 2 2 1 1 − 1 3 1 − 2 − 1 − 2 1 −→ 1 3 2 1 2 − 5 − 3 − 3 − 1 5 3 3 1 −→ 1 3 2 1 2 0 1 3 / 5 3 / 5 1 / 5 0 0 −→ 1 0 1 / 5 − 4 / 5 7 / 5 0 1 3 / 5 3 / 5 1 / 5 0 0 The variables x 3 and x 4 are free, so we can assign them parameters t and s , respectively. Then the solutions are x 1 x 2 x 3 x 4 = − 1 / 5 t + 4 / 5 s + 7 / 5 − 3 / 5 t − 3 / 5 s + 1 / 5 t s = − 1 / 5 − 3 / 5 1 t + 4 / 5 − 3 / 5 1 s + 7 / 5 1 / 5 . 2. (12 points) Consider the following: A = [ 2 1 4 − 1 ] , B = [ 2 3 4 6 ] , C = 2 − 1 3 5 1 , and ⃗v = [ 3 4 ] . Compute each expression, or give a reason why the expression is not defined. (a) A + 2 B Answer. A + 2 B = [ 2 1 4 − 1 ] + 2 [ 2 3 4 6 ] = [ 2 1 4 − 1 ] + [ 4 6 8 12 ] = [ 6 7 12 11 ] ....
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3e-spring2011-exam_1_solutions - Math 3E — Exam 1...

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