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Unformatted text preview: Solutions to Practice Questions for Midterm #2 This “assignment” is not to be handed in and will not be graded. It is intended to help you prepare for the midterm on November 6th. Please note that drop date is November 3rd. Question # 1: Consider the following matrix: A = 3 1 6 6 5 2 4 2 2 4 6 2 2 1 4 2 9 9 1 1 1 2 6 1 2 1 1 3 2 1 1 (a) Verify using row reduction that the RREF of A equals the following matrix R . Show your work. R = 1 2 3 1 1 3 2 1 1 (b) Give the rank of A . (c) Give a basis for Col ( A ) and state its dimension. (d) Give a basis for Row ( A ) and state its dimension. (e) Give a basis for Null ( A ) and state its dimension. (f) Exactly one of the answers to (b)(e) is different if we replace “ A ” with “ R ”. Which one and why? Solution: (a) If you are having difficulties with following the Gaussian algorithm — which is a methodical, careful procedure that leads you to the correct solution with the least amount of effort or extra thought required (and hence, presumably, the fewest possible mistakes) — then you need to sit with someone who can walk you through some examples (eg: me, or someone at the dropin center). (b) Since the RREF of A is R and R has 4 leading 1s, the rank of A is 4. (c) Our theorem from class says that the columns of A corresponding to leading 1s in the RREF of A will give a basis for Col ( A ), so our answer is: 3 2 1 1 , 1 2 4 1 1 , 2 2 1 2 1 , 4 2 1 1 . Since this is a basis, dim( Col ( A )) = 4, which is the same as the rank of A (as it should, by the theorem). 1 2 (d) Our theorem from class says that the nonzero rows of the RREF of A form a basis for the rowspace of A . So our answer is { (1 , , 2 , 3 , 1 , , 0) , (0 , 1 , , 3 , 2 , , 0) , (0 , , , , , 1 , 0) , (0 , , , , , , 1) } (or, write the vectors as columns, it doesn’t matter); thus dim( Row ( A )) = 4, which equals the rank of A , as it should by the theorem. (e) To give a basis for Null ( A ) we have to solve the linear system A~x = ~ 0, which implies row reducing [ A  ~ 0]. But since the RHS is just zeros, it stays as zeros, so we already know what this will row reduce to: [ R  ~ 0]. With that in mind, we can use R above to deduce x 1 + 2 x 3 + 3 x 4 + x 5 = 0 x 2 3 x 4 + 2 x 5 = 0 x 6 = 0 x 7 = 0 and x 3 , x 4 and x 5 are free. So we set x 3 = r , x 4 = s , and x 5 = t so give the general solution x 1 = 2 r 3 s t x 2 = 3 s 2 t x 3 = r x 4 = s x 5 = t x 6 = 0 x 7 = 0 with r, s, t ∈ R . Since the general solution (that is, the set of all solutions) to A~x = ~ 0 is precisely the nullspace of A , we write in vector form:...
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This note was uploaded on 09/28/2011 for the course MATH 1341 taught by Professor Daigle during the Winter '10 term at University of Ottawa.
 Winter '10
 DAIGLE

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