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### quiz1_S07 solutions

Course: MAT 202, Spring 2008
School: Princeton
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University Princeton Department of Mathematics MAT 202 Linear Algebra with Applications QUIZ 1 Spring 2007 SOLUTIONS (1) Consider the system x1 + x 2 - x 3 = 1 2x1 + kx2 + x3 = 5 x1 + x2 + kx3 = k + 2 (a) (10 points) For what values of k does this system have exactly one solution? (b) (10 points) Solve this system with k = -1. Solution: (a) We apply Gauss-Jordan elimination to the system written in matrix...

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University Princeton Department of Mathematics MAT 202 Linear Algebra with Applications QUIZ 1 Spring 2007 SOLUTIONS (1) Consider the system x1 + x 2 - x 3 = 1 2x1 + kx2 + x3 = 5 x1 + x2 + kx3 = k + 2 (a) (10 points) For what values of k does this system have exactly one solution? (b) (10 points) Solve this system with k = -1. Solution: (a) We apply Gauss-Jordan elimination to the system written in matrix notation: 3 2 3 2 1 1 -1 1 1 1 -1 1 3 5 4 2 k 3 1 5 54 0 k-2 0 0 k+1 k+1 1 1 k k+2 The system has exactly one solution when the corresponding rref has three leading ones, that is, when k = 2 and k = -1. (b) We replace k by -1 above and continue elimination: 3 3 2 3 2 2 1 0 0 2 1 1 -1 1 1 1 -1 1 4 0 -3 3 3 5 4 0 1 -1 -1 5 4 0 1 -1 -1 5 0 0 0 0 0 0 0 0 0 0 0 0 x1 = 2 The solutions are the triples (x1 , x2 , x3 ) satisfying x2 - x3 = -1 . Taking x3 as a free variable, the solutions form the vectors 3 3 2 3 2 2 0 2 x1 4 x2 5 = 4 -1 5 + t 4 1 5 , tR. 1 0 x3 (2) (20 points) Let 1 -2 3 v1 = 2 , v2 = -5 , and b = 4 . 3 2 7 Is the vector b a linear combination of v1 and v2 ? Justify your answer. Solution: The vector b is a linear combination of v1 and v2 if there are numbers x and y such that b = xv1 + yv2 , i.e., when there is at least one solution to 2 3 2 3 1 -2 3 x 4 2 -5 5 =4 4 5 . y 3 2 7 0 1 0 3 7 2 5 -18 We solve this by Gauss-Jordan elimination: 3 2 3 2 2 1 1 -2 3 1 -2 3 4 2 -5 4 5 4 0 -1 -2 5 4 0 0 0 8 -2 3 2 7 The last line shows that the system is inconsistent, hence b is not a linear combination of v1 and v2 . 2 (3) (20 points) Find the inverse of the matrix 1 2 1 A = 1 3 2 2 2 -1 Solution: By 2 1 4 1 2 2 1 4 G-J elimination (where the omitted entries 3 2 1 2 1 2 1 1 54 1 1 1 3 2 1 -2 -3 2 -1 3 2 1 -2 -1 3 54 1 1 1 1 -1 1 2 1 -1 -4 we conclude that 2 3 7 -4 -1 -1 3 1 5 . A = 4-5 4 -2 -1 are zero), 3 1 -1 1 5 -2 1 3 7 -4 -1 -5 3 1 5 4 -2 -1 (4) Let L1 be the line in R2 with equation x - y = 0 and L2 the line with equation x y + = 0. Consider the linear transformation T : R2 R2 given by first projecting onto L1 , then rotating counterclockwise by an angle and finally reflecting with respect to L2 . 4 (a) (15 points) What is the matrix of the transformation T ? (b) (5 points) Is this transformation invertible? y L2 PSfrag replacements x L1 Solution: (a) The columns of the matrix of T are T (e1 ) and T (e2 ). The described sequence of transformations has the following effect on the standard vectors: e1 e2 - - 1 e 2 1 1 e 2 1 1 + 2 e2 1 + 2 e2 - - 2 2 2 e 2 2 2 e 2 2 - - - 2 e1 = T (e1 ) 2 - 22 e1 = T (e2 ) Therefore, the matrix is - - 2 2 0 0 . (b) This transformation is not invertible because, in particular, both standard vectors e1 and e2 are mapped to the same vector. 3 (5) Determine whether the following statements are true or false, explaining why. (a) (5 points) If a linear system Ax = b has infinite solutions, then Ax = 0 also has infinite solutions. Solution: TRUE If Ax1 = b and Ax2 = b, then A(x2 - x1 ) = 0. Hence, fixing one solution x1 to the system Ax = b, any other solution x2 produces a solution to the homogeneous equation by taking the difference x2 - x1 . (b) (5 points) There is a linear transformation T : R2 R2 such that T Solution: TRUE Take T with 1 T 0 T 1 2 = 1 1 and T 2 1 = 1 2 . = T ,, -1 2 1 1 2 and = - = - 1 1 2 1 1 ,, 3 -1 2 1 1 0 = - = - = T 0 1 1 0 1 1 -1 3 which has matrix . 1 0 1 2 2 1 Alternatively, the matrix A of T would have to satisfy A = . Since 1 1 1 2 -1 1 2 2 1 1 2 -1 3 is invertible, there is a solution A = = . 1 1 1 2 1 1 1 0 (c) (5 points) If A and B are matrices with AB = B and B = 0, then A must be the identity matrix. 1 0 For instance, A = B = (corresponding to horizontal projection) satisfy 0 0 AB = B and B = 0, but A is not the identity. Solution: FALSE (d) (5 points) If an n n matrix A has A2 = 0, then the matrix In + A is invertible. Solution: TRUE The inverse matrix is In - A because (In + A)(In - A) = In - A + A - A2 = In .
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