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Unformatted text preview: Exam 1 — Version 1 – Solutions Math 340L, Spring 2007, Purcell 1. (5 points) Let A be the matrix: 1 3 6 2 6 12 0 4 1 Find properties of b = b 1 b 2 b 3 such that b is in the span of the columns of A . Show all work. Solution: We row reduce the augmented matrix: 1 3 6 b 1 2 6 12 b 2 0 4 1 b 3 Add 2 times the first row to the second to cancel the 2 in the first column, second row. 1 3 6 b 1 0 0 0 2 b 1 + b 2 0 4 1 b 3 Swap the second and third rows. 1 3 6 b 1 0 4 1 b 3 0 0 0 2 b 1 + b 2 Note there is a pivot in the first and second columns. The vector b will be in the span of the columns of A if and only if there is no pivot in the last column, or iff 2 b 1 + b 2 = 0 . 1 2. The problems below concern the matrix: A = bracketleftbigg 1 2 6 0 2 6 bracketrightbigg (a) (3 points) Describe all solutions to A x = in parametric vector form. Show your work. Solution: We row reduce A . A = bracketleftbigg 1 2 6 0 2 6 bracketrightbigg ∼ bracketleftbigg 1 0 0 0 2 6 bracketrightbigg ∼ bracketleftbigg 1 0 0 0 1 3 bracketrightbigg Now the rows of the reduced matrix give the equations: x 1 = 0 , x 2 + 3 x 3 = 0 , and x 3 is free. In parametric vector form: x 1 x 2 x 3 = x 3  3 1 (b) (2 points) Does A x = b have at least one solution for every possible b ? Why or?...
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This note was uploaded on 03/20/2008 for the course M 340L taught by Professor Pavlovic during the Spring '08 term at University of Texas.
 Spring '08
 PAVLOVIC
 Math, Matrices

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