Math416_HW2_sol - Math 416 - Abstract Linear Algebra Fall...

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Unformatted text preview: Math 416 - Abstract Linear Algebra Fall 2011, section E1 Homework 2 solutions Section 1.3 3.1. (1 pt check) a. 1 2 3 4 5 6 1 3 2 = 13 31 . b. 1 2 0 1 2 0 1 3 = 7 3 2 . c. 1 2 0 0 0 1 2 0 0 0 1 2 0 0 0 1 1 2 3 4 = 5 8 11 4 . d. The dimensions dont match (3 6 = 4). 3.2. Let T : R 2 R 2 denote the transformation in question. Its matrix is A = T (- e 1 ) T (- e 2 ) = 0 1 1 0 . 3.3. (2 pts) a. T x y = x + 2 y 2 x- 5 y 7 y = 1 2 2- 5 7 x y says the matrix of T is A = 1 2 2- 5 7 . c. T (1) = 0, T ( t ) = 1, T ( t 2 ) = 2 t , ... , T ( t n ) = nt n- 1 . The matrix of T is 0 1 0 ... 0 0 2 ... 0 0 0 ... . . . . . . . . . . . . . . . 0 0 0 ... n 0 0 0 ... ....
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Math416_HW2_sol - Math 416 - Abstract Linear Algebra Fall...

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