midterm_S06_solutions

# midterm_S06_solutions - MAT 202 MIDTERM SPRING 2006(1 Let T...

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MAT 202 MIDTERM SPRING 2006 (1) Let T : R 2 ! R 2 be a linear transformation such that T ° 2 3 ± = ° 1 1 ± ; T ° 1 2 ± = ° 1 ° 1 ± : (a) (8 points) Find T ( °! e 1 ) and T ( °! e 2 ) . Solutions: Assume T ( °! x ) = A °! x , then A ° 2 1 3 2 ± = ° 1 1 1 ° 1 ± ; hence A = ° 1 1 1 ° 1 ± ° 2 1 3 2 ± ° 1 = ° ° 1 1 5 ° 3 ± : This shows T ( °! e 1 ) = ° ° 1 5 ± ; T ( °! e 2 ) = ° 1 ° 3 ± : (b) (7 points) Find the matrix of the composition R ± T where R : R 2 ! R 2 is the counterclockwise rotation by 90 o . Solutions: The matrix is given by ° 0 ° 1 1 0 ± ° ° 1 1 5 ° 3 ± = ° ° 5 3 ° 1 1 ± : (2) Let A = 2 6 6 4 3 1 3 4 1 0 2 1 0 1 ° 3 3 2 0 4 3 3 7 7 5 : (a) (5 points) Find the reduced row echelon form of A . Solutions: rref A = 2 6 6 4 1 0 2 0 0 1 ° 3 0 0 0 0 1 0 0 0 0 3 7 7 5 : (b) (5 points) Find a basis for im A . Solutions: A basis is 2 6 6 4 3 1 0 2 3 7 7 5 ; 2 6 6 4 1 0 1 0 3 7 7 5 ; 2 6 6 4 4 1 3 3 3 7 7 5 : 1

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2 MAT 202 MIDTERM SPRING 2006 (c) (5 points) Find a basis for ker A . Solutions: The solution to A °! x = °! 0 is given by °! x = 2 6 6 4 ° 2 s 3 s s 0 3 7 7 5 = s 2 6 6 4 ° 2 3 1 0 3 7 7 5 : Hence a basis of ker A may be chosen as 2 6 6 4 ° 2 3 1 0 3 7 7 5 .
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