00aut-m2sols

00aut-m2sols - MATH 51 MIDTERM 2 SOLUTIONS 1. (a) Compute...

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MATH 51 MIDTERM 2 SOLUTIONS 1. (a) Compute the inverse of the matrix 13 - 2 02 4 00 - 1 Solution. - 2 - 1 ± ± ± ± ± ± 100 010 001 - 2 01 2 - 1 ± ± ± ± ± ± 01 / 20 10 - 8 - 1 ± ± ± ± ± ± 1 - 3 / / - 8 00 1 ± ± ± ± ± ± 1 - 3 / / - 1 ± ± ± ± ± ± 1 - 3 / 2 - 8 / 22 - 1 so A - 1 = 1 - 3 / 2 - 8 / - 1 (b) For which value(s) of x is the matrix below not invertible? Explain your answer. 111 012 5 x 6 Solution. The determinant of this matrix is 1(6 - 2 x ) - 1(0 - 10)+1(0 - 5) = 11 - 2 x A matrix is not invertible if and only if its determinant equals zero, so this matrix is not invertible if and only if x =11 / 2. 2. (a) Suppose A = 2 / 3 - 1 / 32 / 3 2 / / 3 - 1 / 3 - 1 / / / 3 1
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is the matrix ofa linear transformation which is geometrically a 60degree rotation about a line L in R 3 . Find the matrix of a 120 degree rotation about L . Hint: Think about composition. Solution. A 120 degree rotation is accomplished by composing the linear trans- formation with itself. Since the matrix for a composition of two transformations is the product of the matrices for the two transformations, the matrix for this rotation is A 2 = 001 100 010 (b) Let B = 2235 4321 - 12 - 9858 v = 6 4 - 2 10 Compute B - 1 v . Hint: You do not need to compute B - 1 .C om p a r e v with the columns of B . Solution. The vector v is twice the third column of B .S i n c e B e 3 equals the third column of B , multiplying by 2 gives v =2 B e 3 = B (2 e 3 ) and therefore B - 1 v e 3 = 0 0 2 0 3. Let Δ 1 be the triangle with vertices (0 , 0), ( - 1 , 0) and (0 , 2) and let Δ 2 be the triangle with vertices (0 , 0), (2 , 0) and (3 , 3). Suppose T : R 2 R 2 is a linear transformation such that T 1 )=Δ 2 .
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00aut-m2sols - MATH 51 MIDTERM 2 SOLUTIONS 1. (a) Compute...

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