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Unformatted text preview: Math 51 Exam 2 Solutions — May 20, 2008 1. (8 points) Compute the following determinant: 2 4 2 1 1 1 1 3 2 1 1 2 2 3 1 Due to the lack of many zeros, it’s convenient to begin with a few row operations: 2 4 2 1 1 1 1 3 2 1 1 2 2 3 1 swap swap = 1 1 1 2 4 2 1 3 2 1 1 2 2 3 1 2 I 3 I 2 I = 1 1 1 2 1 1 2 1 1 1 +2 III = 1 1 1 4 3 1 2 1 1 1 · ( 1) = 1 1 1 0 0 4 3 0 1 2 1 0 0 1 1 [expand along first column] = (1) 4 3 1 2 1 1 1 [expand along first column] = (1) (1) 4 3 1 1 = (4 · 1 ( 3)( 1)) = 1 Math 51, Spring 2008 Exam 2 Solutions — May 20, 2008 Page 2 of 10 2. (15 points) Let B be the basis { v 1 , v 2 , v 3 } of R 3 where v 1 = 1 1 1 , v 2 =  1 1 , v 3 =  1 1 . (You don’t have to check that B is a basis.) Let T : R 3 → R 3 be the linear transformation defined by the following formulas: T ( v 1 ) = v 2 , T ( v 2 ) = v 3 , T ( v 3 ) = v 1 . (a) Find the matrix of T with respect to the basis B . (4 points) Write B for the matrix of T with respect to B . Recall that the i th column of B is [ T ( v i )] B , i.e., the coordinates of the vector T ( v i ) with respect to B . Thus, B = [ T ( v 1 )] B [ T ( v 2 )] B [ T ( v 3 )] B = [ v 2 ] B [ v 3 ] B [ v 1 ] B = 0 0 1 1 0 0 0 1 0 . Notes: The most common mistake was to forget to express the outputs of T in coordinates with respect to the basis B ; thus, many people instead used the standard coordinates of the v i vectors as the columns for their matrix. (b) Find the matrix of T with respect to the standard basis. (7 points) Note that we seek the matrix A satisfying T ( x ) = A x for all x in R 3 . Solution 1: The technique most people used was to use the changeofbasis formula A = CBC 1 , where C is the matrix whose i th column is v i . This technique is correct, but it is easy to make mistakes when computing the inverse of a matrix, and the operations involved take a lot of time. (Also, some people wrote A = C 1 BC , which is incorrect.) For completeness, here is a summary of the computations involved: if C is the changeofbasis matrix for B , then C = 1 1 1 1 1 1 1 , and we must compute (via row reduction) that C 1 = 1 3 1 3 1 3 2 3 1 3 1 3 1 3 2 3 1 3 . Thus, A = CBC 1 = 1 1 1 1 1 1 1 0 0 1 1 0 0 0 1 0 1 3 1 3 1 3 2 3 1 3 1 3 1 3 2 3 1 3 =  1 1 1 1 1 1 1 1 3 1 3 1 3 2 3 1 3 1 3 1 3 2 3 1 3 =  1 0 1 1 0 1 . Math 51, Spring 2008 Exam 2 Solutions — May 20, 2008 Page 3 of 10 Solution 2: A more efficient approach is to write the standard basis vectors in terms of the vectors in B . If we inspect these vectors, we can see that 3 e 3 = v 1 + v 2 + v 3 ; applying T to both sides (and using linearity) we get T (3 e 3 ) = T ( v 1 + v 2 + v 3 ) = T ( v 1 )+ T ( v 2 )+ T ( v 3 ) = v 2 + v 3 + v...
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 Linear Algebra, Algebra, Determinant, Differential Calculus, Orthogonal matrix, Standard basis

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