mid1w05s

# mid1w05s - 1-2 4 1-4 8 2-→ 1-1 2 2 3-6-3 6-→ 1-1 2 2...

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Math 20F First Exam Solutions 26 January 2005 1. (a) Yes, because x 2 and x 3 are free variables. (b) No, because there is a row of zeros. (When the augmented matrix is put in row echelon form, the last column can be anything, depending on the choice of b and so the last equation will become “zero = anything”. 2. (a) Unde±ned: A is 2 × 3 but A T is not. (b) Unde±ned: for BC to be de±ned, the number of rows of C must equal the number of columns of B . With A = B = C , this is not true. (c) b 5 - 2 - 2 2 B . (d) Unde±ned: To have an inverse, a matrix must have the same number of rows and columns. 3. The augmented matrix and reduction to row echelon form: 1 - 1 2 2 2
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Unformatted text preview: 1-2 4 1-4 8 2 -→ 1-1 2 2 3-6-3 6 -→ 1-1 2 2 3-6 Thus x 3 is a free variable. The second row tells us that x 2 = 2 x 3 . The ±rst row tells us that x 1 = 2 + x 2-2 x 3 = 2. Thus we have x 1 = 2 x 2 = 2 x 3 x 3 free . 4. (a) all p ≥ 4 (b) all p ≤ 4 5. Since the number of columns of A T equals the number of rows of A , the product is de±ned. Since A T has p rows and A has p columns, A T A is p × p . Recalling that ( BC ) T = C T B T and ( B T ) T = B , we have ( A T A ) T = A T ( A T ) T = A T A....
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