assign9_soln

assign9_soln - Math 136 Assignment 9 Solutions 1. Calculate...

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Unformatted text preview: Math 136 Assignment 9 Solutions 1. Calculate the area of the parallelogram determined by the following vectors. a) 2 −5 , 3 2 Solution: A = det 2 −5 32 = |2(2) − (−5)(3)| = 19 1 −3 b) 0 , 1 −5 1 Solution: From our work in deriving the formula for the volume of a parallelepiped, we found that the area of the parallelogram induced by u, v ∈ R3 is A= u×v Hence, we get 5 −3 1 √ A = 0 × 1 = 14 = 222 1 1 −5 2. Find the volume of the parallelepiped determined by the following vectors 5 1 1 −1, 4 , 2 a) 1 −1 −6 0 6 31 5 1 1 2 = det 0 3 −4 = −117 Solution: V = det −1 4 1 −1 6 1 −1 −6 1 1, b) 4 1 3, 4 −2 1 −5 1 1 −2 1 1 −2 Solution: V = det 1 3 1 = det 0 2 3 = 6 4 4 −5 00 3 1 2 3. Use Cramer’s Rule to solve the following systems. a) x1 − 5x2 − 2x3 = −2 2x1 + 3 x3 = 3 4x1 + x2 − x3 = 1 1 −5 −2 3 , so det A = −77 and Solution: The coefficient matrix is A = 2 0 4 1 −1 x1 = x2 = 1 −2 −2 2 1 23 3= −77 4 1 −1 11 x2 = x1 30/77 Hence, x2 = 2/11 . 57/77 x3 −2 −5 −2 1 30 3 0 3= −77 1 77 1 −1 1 −5 −2 57 1 20 3= −77 4 1 77 1 3 b) x1 + 2 x3 3x1 − x2 + 3x3 −2x1 + x2 1 0 Solution: The coefficient matrix is A = 3 −1 −2 1 = −2 =5 =0 2 3, so det A = −1 and 0 −2 0 2 1 5 −1 3 = −16 x1 = −1 0 10 1 −2 2 1 3 5 3 = −32 x2 = −1 −2 0 0 x2 = 1 0 −2 1 3 −1 5 = 7 −1 −2 1 0 x1 −16 Therefore, x2 = −32. 7 x3 2 −1 0 1 0 −1 3 2 −1 −1 4. Let A = 0 1 0 0. Use the cofactor method to calculate (A )23 and (A )42 . 0 2 03 −1 3 2 32 = −18. So We have det A = 2 1 0 0 = −2 03 2 03 −1 201 1 18 2+3 0 3 2= = (−1) =1 −18 18 003 −1 2 −1 0 1 2+4 0 1 0 =0 = (−1) −18 020 (A )23 (A )42 ...
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This note was uploaded on 07/13/2011 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.

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assign9_soln - Math 136 Assignment 9 Solutions 1. Calculate...

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