Calculus 2 - Assignment 4 Solutions

# Calculus 2 - Assignment 4 Solutions - MAT1332 Spring/Summer...

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Unformatted text preview: MAT1332 Spring/Summer 2009 Assignment 4 Solutions 1. First observe that the first equation and the third are the same. Thus, we are simply work- ing with two equations 2 x + 3 y + v = and 2 y + z = in four unknowns. To obtain the non-trivial solutions we observe that the coefficient ma- trix of the system 2 3 0 1 0 2 1 0 is already in row-echelon form and the free variables are z and v . First, we set z : = 2, then the second equation yields y =- 1. Setting v : = 1 in the first equation yields 2 x- 3 + 1 = and therefore x = 1. 2. The expressions evaluate to (a) - 4- 25- 2- 5- 15 8- 5 6 8 (b) - 63 67- 75 11 (c) - 54- 150- 4- 9 66 63 (d) - 12 44 9 5- 42- 36 (e) 38 20- 1- 29- 30 31 68- 8 68 (f) - 9 1- 6 11 2- 3 (g) 351- 201 461 205 1 3. We test the determinants to find det( A ) = 1, det( B ) = 0 and det( C ) = - 1. Thus, A and C are invertible, but B is not. To determine the inverses we augment the matrices and apply the elementary row operations as follows: 6 1 4 1 0 0 1- 2 3 0 1 0 2 8- 7 0 0 1 R 3- 2 * R 2 / 6 1 4 1 1- 2 3 1 12- 13- 2 1 R 2- 1 6 R 1 / 6 1 4 1- 13 6 7 3- 1 6 1 12- 13- 2 1 R 3 + 72 13 R 2 / 6 1 4 1- 13 6 7 3- 1 6 1- 1 13- 12 13 46 13 1 R 2 + 91 3 * R 3 / 6 1 4 1- 13 6-...
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## This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.

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Calculus 2 - Assignment 4 Solutions - MAT1332 Spring/Summer...

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