232sol9

# The coefcients x y and z that we want to use to

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Unformatted text preview: are 10, 3, and 8, so e1 = 10v1 3v2 + 8v3 . Similarly, e2 = 4v1 + v2 Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 3v3 , e3 = 3e1 + e2 (d) The row reduction that we did in step (c) is exactly what you do to calculate A off the 2 matrix on the right hand side after we row reduce A to the identity matrix. So 3 10 4 3 1 1 5. A 1=4 3 8 3 2 1. 2e 3 . Just read (e) The coefﬁcients for the linear combinations in part (c) are found by reading the columns of A 1. A2. (a) By the properties of linear transformations: T (e1 ) = T ( 10v1 3v2 + 8v3 ) = 10T (v1 ) 3T ( v 2 ) + 8 T ( v 3 ) = 10(1, 1) 3(2, 5) + 8(0, 1) = ( 16, 3). (b) By the same method as we used for part (a), we can calculate T (e2 ) = (6, 2) and T (e3 ) = (5, 0),...
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## This document was uploaded on 03/03/2014 for the course MATH 232 at Simon Fraser.

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