term_test_1_w10_soln

# term_test_1_w10_soln - Math 136 Term Test 1 Solutions 1...

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Unformatted text preview: Math 136 Term Test 1 Solutions 1. Short Answer Problems [2] a) Calculate proj (1 , 1 , 2) (- 1 , 2 , 2). Solution: proj (1 , 1 , 2) (- 1 , 2 , 2) = (- 1 , 2 , 2) Â· (1 , 1 , 2) k (1 , 1 , 2) k 2 (1 , 1 , 2) = 5 6 (1 , 1 , 2). [1] b) If ~n = ~a Ã— ~ b , then what is ~a Â· ~n ? Solution: Since ~a Ã— ~ b gives an vector orthogonal to ~a , we have that ~a Â· ~n = 0. [1] c) What is the definition of the rank of a matrix? Solution: The number of leading 1s (pivot positions) in the RREF of the matrix. [2] d) Let A = 1 2- 1 0 3 4 and B = - 6 5 2 1 1 . Calculate AB . Solution: AB = 9- 3 19 10 [2] e) Find the area of the parallelogram induced by ~a = (1 , 2 ,- 1) and ~ b = (3 , 2 ,- 1). Solution: Area= k (1 , 2 ,- 1) Ã— (3 , 2 ,- 1) k = k (0 ,- 2 ,- 4) k = âˆš 20. [3] f) Let S = { ~v 1 ,~v 2 ,~v 3 } be a set of three different vectors in R 3 . What are the possible geometric configurations of span S ? Solution: If all three are linearly independent we get span S = R 3 . If ~v 3 = ~v 1 + ~v 2 and ~v 1 ,~v 2 are linearly independent, then span S is a plane. If ~v 1 = 2 ~v 2 = 3 ~v 3 , then span S is a line. Note that since all three vectors must be different, we can not have span S = { ~ } . 1 2 2. Consider the system of linear equations: 2 x 3 + 6 x 4 =- x 1 + 2 x 2 + 3 x 3 + x 4 = 5 x 1- x 2- 4 x 3- x 4 = 1 [1] a) What is the augmented matrix of the system? Solution: 2 6- 1 2 3 1 5 1- 1- 4- 1 1 . [4] b) Row reduce the augmented matrix of the system into reduced row echelon form, stating the elementary row operations used. Solution: 2 6- 1 2 3 1 5 1- 1- 4- 1 1 R 1 â†” R 3 1- 1- 4- 1 1- 1 2 3 1 5 2 6 R 2 + R 1 1 2 R 3 1- 1- 4- 1 1 1- 1 6 1 3 R 1 + 4 R 3 R 2 + R 3 1- 1 0 11 1 1 3 6 1 3 R 1 + R 2...
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term_test_1_w10_soln - Math 136 Term Test 1 Solutions 1...

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