term_test_1_w10_soln

term_test_1_w10_soln - Math 136 1. Short Answer Problems...

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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) · (1 , 1 , 2) ± (1 , 1 , 2) ± 2 (1 , 1 , 2) = 5 6 (1 , 1 , 2). [1] b) If ±n = ±a × ± b , then what is · ? Solution: Since × ± b gives an vector orthogonal to , we have that · = 0. [1] c) What is the deFnition of the rank of a matrix? Solution: The number of leading 1s (pivot positions) in the RRE± of the matrix. [2] d) Let A = ± 12 - 1 0 3 4 ² and B = 0 - 6 52 11 . Calculate AB . Solution: AB = ± 9 - 3 19 10 ² [2] e) ±ind the area of the parallelogram induced by = (1 , 2 , - 1) and ± b = (3 , 2 , - 1). Solution: Area= ± (1 , 2 , - 1) × (3 , 2 , - 1) ± = ± (0 , - 2 , - 4) ± = 20. [3] f) Let S = { ±v 1 ,±v 2 3 } be a set of three di²erent vectors in R 3 . What are the possible geometric conFgurations of span S ? Solution: If all three are linearly independent we get span S = R 3 . If 3 = 1 + 2 and 1 2 are linearly independent, then span S is a plane. If 1 =2 2 =3 3 , then span S is a line. Note that since all three vectors must be di²erent, we can not have span S = { ± 0 } . 1
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2 2. Consider the system of linear equations: 2 x 3 +6 x 4 =0 - 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: 0 0 2 6 0 - 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: 0 0 2 6 0 - 1 2 3 1 5 1 - 1 - 4 - 1 1 R 1 R 3 1 - 1 - 4 - 1 1 - 1 2 3 1 5 0 0 2 6 0 R 2 + R 1 1 2 R 3 1 - 1 - 4 - 1 1 01 - 10 6 0 0 1 3 0 R 1 +4 R 3 R 2 + R 3 1 - 1 0 11 1 0 1 0 3 6 0 0 1 3 0 R 1 + R 2 1 0 0 14 7 0 1 0 3 6 0 0 1 3 0 [2] c) Find the general solution of the system.
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term_test_1_w10_soln - Math 136 1. Short Answer Problems...

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