Calculus 2 - Assignment 3 Solutions

# Calculus 2 - Assignment 3 Solutions - MAT1332 Spring/Summer...

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MAT1332 Spring/Summer 2009 Assignment 3 Solutions 1. We will solve this problem by replacing the system of equations with its coefﬁcient matrix: C = 7 0 - 1 8 4 2 0 2 - 6 7 3 - 10 We operate using the elementary row operations as follows: C R 2 - 4 7 * R 1 / 7 0 - 1 8 0 2 4 7 - 18 7 - 6 7 3 - 10 R 3 + 6 7 * R 1 / 7 0 - 1 8 0 2 4 7 - 18 7 0 7 15 7 - 22 7 R 3 - 7 2 * R 2 / 7 0 - 1 8 0 2 4 7 - 18 7 0 0 1 7 41 7 7 * R 3 / 7 0 - 1 8 0 2 4 7 - 18 7 0 0 1 41 R 1 + R 3 / 7 0 0 49 0 2 4 7 - 18 7 0 0 1 41 R 2 - 4 7 * R 3 / 7 0 0 49 0 2 0 - 26 0 0 1 41 1 7 * R 1, 1 2 * R 2 / 1 0 0 7 0 1 0 - 13 0 0 1 41 Thus the solution is x 1 = 7, x 2 =- 13 and x 3 = 41. 2. By Euler’s formula e i ϑ + e - i ϑ = cos( ϑ ) + i sin( ϑ ) + cos( - ϑ ) + i sin( - ϑ ) = cos( ϑ ) + i sin( ϑ ) + cos( ϑ ) - i sin( ϑ ) = 2cos( ϑ ), where the penultimate equation is by the fact that cos( - ϑ ) = cos( ϑ ) and sin( - ϑ ) =-

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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 3 Solutions - MAT1332 Spring/Summer...

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