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# sol2 - is 1 x x^2 x^3 x^4 x^5 x^6 x^7 Thus the dimension of...

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Solutions to 20F Midterm #2 February 25, 2009 (1) Convert A to echelon form U. The pivot columns of U are columns #1 and #3. Thus columns #1 and #3 of A form a basis for CS(A). (2) True. The determinant would be nonzero if and only if A had 4 (nonzero) pivots, but there are only 2 pivots, so the determinant of A is zero. (3) True. Since CS(A) is a line, CS(A) has dimension 1, so there must be exactly one pivot column. In computing NS(A), since there is only one pivot variable, the other two variables must be free variables. Thus the null space NS(A) has dimension 2, so it's a plane. Finally, the plane goes through the origin, because all vector spaces, including NS(A), must contain the zero vector. (4) A basis for V
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Unformatted text preview: is 1, x, x^2, x^3, x^4, x^5, x^6, x^7. Thus the dimension of V is 8. (5) The two columns of A are independent. The column vector b is in CS(A) if and only if the 3 by 3 augmented matrix [A|b] has dependent columns, i.e., if and only if the augmented matrix does not have 3 pivots. An echelon form for this augmented matrix has first row 2 1 b1, second row 0 5 b2+2b1, and third row 0 0 b3+(b1-7b2)/5. This last entry must be 0 in order to avoid having 3 pivots, i.e., b3 = (7b2 - b1)/5. (6i) True: T(v2) = 1v1 + 0 v2 = v1. (6ii) True: The matrix 0 2 1 equals B^-1 A B, where A is the matrix 0 1 2 and B is the matrix that takes the old basis to the new, i.e., B is 1 1 1 2 . Note that B^-1 is 2 -1-1 1 ....
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