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**Unformatted text preview: **Math 331 Midterm 2 practice problems This is a list of practice problems for the first midterm in Math 331 (Wylie). The actual exam will be 10 problems long with each problem worth 10 points. The first problem on the exam will be similar to the first practice problem and will be a true/false with 5 parts. You will not need to show any work on this problem. 1. True of False: (a) If A and B are square matrices then det( A + B ) = det( A ) + det( B ). False. (b) If A is a 5 × 7 matrix with rank 5 then there is only one solution to A x = . False. (Nul( A ) is a two dimensional subspace of R 7 by the rank theorem, so there are infinitely many solutions.) (c) Every basis of R 5 has 5 elements in it. True. (d) If A , B , and C are nonzero n × n matrices and AB = AC , then B = C . False. (e) If the columns of B are linearly independent then so are the columns of BA . False. (If A has linearly dependent columns then so will BA no matter what B is.) 2. Determine whether the following functions are linear transformations. If the function is a linear transformation find the standard matrix that represents it and determine whether the linear transformation is one-to-one and whether it is onto. (a) T ( x,y ) = ( x,xy ) Solution: The function is not a linear transformation. For example, T (1 , 0) = (1 , 0) T (0 , 1) = (0 , 0) T (1 , 1) = (1 , 1) so T u + T v 6 = T ( u + v ). (b) T ( x,y,z ) = (2 x + 6 y- z, 3 x- 7 y ) Solution: The function is a linear transformation since T ( x,y,z ) = (2 x + 6 y- z, 3 x- 7 y ) = x 2 3 + y 6- 7 + z- 1 = 2 6- 1 3- 7 x y z So the standard matrix of T is A = 2 6- 1 3- 7 The transformation is one to one if there are no nontrivial solutions to A x = . In other words, if every column contains a pivot. Since A is a 2 × 3 matrix, it is impossible for every column to contain a pivot. So T is not one to one. The transformation is onto if there is a solutions to A x = b for every b . In other words, if every row contains a pivot. Row reducing A gives us 2 6- 1 3- 7 II- 3 2 I----→ 2 6- 1- 16 3 2 Since both rows of A contain a pivot, T is onto. (c) T ( x,y,z ) = ( x + 2 y- 4 z- 6 ,- x + 5 y + 9 ,z ) Solution: The function is not a linear transformation since T (0 , , 0) = (- 6 , 9 , 0) and for linear transformations T ( ) = always. (d) T : R 2 → R 2 is the rotation of the plane by π 4 in counterclockwise direction. Solution: Rotations are linear transformations, so we need to find the standard matrix of A . The columns of A are T ( e 1 ) and T ( e 2 ), T ( e 1 ) is the vector e 1 = 1 rotated by π 4- counterclockwise is √ 2 2 √ 2 2 . T ( e 2 ) is the vector e 2 = 1 rotated by π 4-counterclockwise is - √ 2 2 √ 2 2 . So A is the matrix A = √ 2 2 √ 2 2 √ 2 2- √ 2 2 The determinant of A is 1, so A is invertible. In particular A is both one to one and onto....

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