Quiz2Solutions

Quiz2Solutions - Newberger Math 247 Spring 02 Solutions for...

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Newberger Math 247 Spring 02 Solutions for Quiz 1.3 and 1.4 Let b = 2 - 1 6 u 1 = 1 - 2 0 u 2 = 0 1 2 u 3 = 5 - 6 8 . 1. (12 points) Show that b is in Span { u 1 , u 2 , u 3 } . In other words show that b is a linear combination of u 1 , u 2 and u 3 . We want to show we can find numbers x 1 , x 2 and x 3 , such that x 1 u 1 + x 2 u 2 + x 3 u 3 = b . To see this we reduce the following augmented matrix to echelon form: 1 0 5 2 - 2 1 - 6 1 0 2 8 6 1 0 5 2 0 1 4 3 0 2 8 6 1 0 5 2 0 1 4 3 0 0 0 0 . This system is consistent, so b is in Span { u 1 , u 2 , u 3 } . 2. (4 points) Name three real numbers, x 1 , x 2 and x 3 , such that b = x 1 u 1 + x 2 u 2 + x 3 u 3 . In (1), we showed the numbers x 1 , x 2 and x 3 exist; in this problem we are asked to find such numbers. To solve for x 1 , x 2 and x 3 , put the matrix from (1) in reduced echelon form (as it turns out, in this example it is already in reduced echelon form! 1 0 5
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This note was uploaded on 01/12/2010 for the course MATH 247 taught by Professor F,newberger during the Spring '03 term at Stanford.

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Quiz2Solutions - Newberger Math 247 Spring 02 Solutions for...

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