136_tt1_f10_soln

# 136_tt1_f10_soln - Math 136 1 Short Answer Problems Term...

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Math 136 Term Test 1 Solutions 1. Short Answer Problems [3] a) State the three types of elementary row operations. Solution: (1) Add a multiple of one row to another. (2) Multiple a row by a non-zero scalar. (3) Swap two rows [2] b) Let ~x = 1 2 1 . Find a unit vector that is orthogonal to ~x . Justify your answer. Solution: There are many possible answers. One possible answers is ~v = 1 2 - 1 0 1 , since ~x · ~v = - 1 2 + 0 + 1 2 = 0 and ~v · ~v = 1 2 + 0 + 1 2 = 1 [2] c) Calculate 3 2 1 3 - 1 - 1 ± 1 - 1 - 2 1 ² . Solution: 3 2 1 3 - 1 - 1 ± 1 - 1 - 2 1 ² = - 1 - 1 - 5 2 1 0 [1] d) Is 1 0 0 , 0 1 0 , 0 0 1 , 1 4 - 3 a basis for R 3 ? Explain. Solution: This cannot be a basis for R 3 since it has 4 vectors. In particular, the set must be linearly dependent. [2] e) Let ~a, ~ b,~ c R 3 such that ~ b × ~ c 6 = ~ 0. Use a geometrical argument to explain why ~a × ( ~ b × ~ c ) = s ~ b + t~ c for some s, t R . Solution: Suppose that ~n = ~ b × ~ c . Then ~n is orthogonal to both ~ b and ~ c , so it is a normal vector to the plane through the origin that contain ~ b and ~ c . Then ~a × ( ~ b × ~ c ) = ~a × ~n is orthogonal to ~n so it lies in the plane with normal ~n , that is, in the plane containing ~ b and ~ c . Hence, for some real numbers s, t , a × ( ~ b × ~ c ) = s ~ b + t~ c since ~ b and ~ c span the plane. 1

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2 2. Consider the system of linear equations: x 1 + x 2 + x 3 = 1 x 1 + 2 x 2 + 2 x 3 = 0 2 x 1 + x 2 + x 3 = 3 [1] a) What is the augmented matrix of the system? Solution: 1 1 1 1 1 2 2 0 2 1 1 3 [2] b) Row reduce the augmented matrix of the system into reduced row echelon form, stating the elementary row operations used. Solution: 1 1 1 1 1 2 2 0 2 1 1 3 R 2 - R 1 R 3 - 2 R 1 1 1 1 1 0 1 1 - 1 0 - 1 - 1 1 R 1 - R 2 R 3 + R 2 1 0 0 2 0 1 1 - 1 0 0 0 0 [2] c) Find the general solution of the system. Solution: We have that x 3 is a free variable, so let x 3 = t R . Then, we get general solution ~x = x 1 x 2 x 3 = 2 - 1 - t t = 2 - 1 0 + t 0 - 1 1
3 3. Let B = 1

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## This note was uploaded on 11/25/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.

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136_tt1_f10_soln - Math 136 1 Short Answer Problems Term...

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