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06aut-m1sols

# 06aut-m1sols - Math 51 Midterm I Solutions(These solutions...

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Math 51 Midterm I Solutions (These ’solutions’ are written more in the form of a guide to correct think- ing, and a guide to the level of understanding we were trying to test. This is for students who want to learn from their mistakes and improve their understanding.) PROBLEM 1(a). This is a routine row reduced echelon form problem. It tests a computational skill only. Here is the answer. 1 0 1 0 1 0 1 2 0 3 0 0 0 1 1 0 0 0 0 0 If you missed it, just check your arithmetic and see where you slipped up. PROBLEM 1(b). The point here is that even though we don’t know the entries a and b yet, we can start the rref process and get a lot of information. Remember, the final rref form will tell us the dimension of the null space, which will be the number of free variables. 1 1 1 1 2 a 1 3 b 1 1 1 0 1 a - 1 0 2 b - 1 1 1 1 0 1 a - 1 0 0 1 + b - 2 a At this point you see that rref( A ) will definitely have pivots in the first two rows and columns. One of two things happens after that. If 1+ b - 2 a = 0 then the rref will have three pivots in the end and N ( A ) = { 0 } . If 1 + b - 2 a = 0 then the last row of rref( A ) will be all 0’s. There is then one free variable and the null space is a line in R 3 . PROBLEM 2(i). There are two basic views of a matrix-vector product A v . The first view is tested here. Multiply the respective columns of A by the entries of v and add. So here A v = 1 a 1 + 2 a 2 + 3 a 3 + 4 a 4 (See the bottom of page 47 of the text.) 1

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Notice this view makes it crystal clear that vectors of the form A v are exactly the vectors in the column space of A , that is, in the span of the columns of A PROBLEM 2(ii). This is the second important view of a matrix-vector prod- uct. The entries of A v are obtained as dot products of the rows of A and the vector v . Let’s ignore the business with transposes and just treat a row as a vector, even though it is written sideways, not up and down. Then A v = r 1 v r 2 v r 3 v . (See the bottom of page 48 of the text.) Notice this view makes it crystal clear that vectors v in the null space of A are exactly the vectors that are orthogonal to all rows of A . PROBLEM 2(iii). The answer is ‘no’. Any four vectors in R 3 are linearly dependent because if you put those vectors in the columns of a 3 by 4 matrix, like A here, then rref( A ) has at most three pivots, therefore there are free variables and hence non-zero vectors in N ( A ). But any non-zero vector in N ( A ) gives rise to a linear dependence relation between the columns of A , using the formula for A v as a linear combination of columns of A . PROBLEM 2(iv). Infinitely many. The explanation just above in 2(iii) explains why the null space of A contains at least a line in R 4 . There is at least one free variable, maybe more.
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06aut-m1sols - Math 51 Midterm I Solutions(These solutions...

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