Spring 2005 - Nagy's Class - Quiz 2

Spring 2005 - Nagy's Class - Quiz 2 - -1 and rotating 1 by...

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Math 20F Quiz 2 (version 1) April 22, 2005 1. (1.7.14) Find the value of h for which the set of vectors 1 - 1 - 2 , - 5 6 11 , 2 - 1 h is linearly dependent . Justify your answer. 1 - 1 - 2 , - 5 6 11 , 2 - 1 h will be linearly dependent if and only if the homogeneous vector equation x 1 1 - 1 - 2 + x 2 - 5 6 11 + x 3 2 - 1 h = 0 0 0 has a nontrivial solution. The augmented matrix of the corresponding homogeneous matrix equation is 1 - 5 2 0 - 1 6 - 1 0 - 2 11 h 0 , which is row equivalent to 1 - 5 2 0 0 1 1 0 0 0 3 + h 0 , which has has nontrivial solutions when h = - 3. Thus, the set of vectors is linearly dependent precisely when h = - 3. A typical dependence relation is - 7 1 - 1 - 2 - - 5 6 11 + 2 - 1 - 3 = 0 0 0 . 2. (1.9.3) Find the standard matrix for the linear transformation T : 2 2 that rotates points (about the origin) through 3 π 2 radians (counterclockwise). Rotating ± 1 0 ² by 3 π 2 radians yields
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Unformatted text preview: -1 and rotating 1 by 3 2 radians yields 1 . Thus, T ( e 1 ) = T 1 = -1 T ( e 2 ) = T 1 = 1 Hence, the standard matrix for T is [ T ( e 1 ) T ( e 2 ) ] = 1-1 . 3. (2.2.24) Suppose A is n n and the equation A x = b has a solution for each b in n . Explain why A must be invertible. [ Hint: Is A row equivalent to I n ?] Since A x = b has a solution for each b in n , we can solve each of the n matrix equations A u 1 = e 1 A u 2 = e 2 A u n = e n Thus, A [ u 1 u 2 u n ] = [ A u 1 A u 2 A u n ] = [ e 1 e 2 e n ] = I . This means that the matrix B = [ u 1 u 2 u n ] satisFes AB = I , which implies that A is invertible....
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This note was uploaded on 04/23/2008 for the course MATH 20F taught by Professor Buss during the Spring '03 term at UCSD.

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