Midterm 2 ReviewSolutions

# Midterm 2 ReviewSolutions - Math 5A Midterm 2 Review...

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Math 5A - Midterm 2 Review Problems - Solutions Winter 2009 The exam will focus on topics from Section 3.6 and Chapter 5 of the text, although you may need to know additional material from Chapter 3 (covered in 3C) or from Chapter 4 (covered earlier this quarter). Below is an outline of the key topics and sample problems of the type you may be asked on the test. Many are similar to homework problems you have done–just remember that you will be required to show your work and/or justify your answers on the exam. 3.6: Span, Linear (In)dependence, Basis, Dimension. 1. Determine if each list of vectors is linearly dependent or independent. Justify your answers. (a) (1 , 2) , (2 , 1) Solution. We write c 1 (1 , 2)+ c 2 (2 , 1) = (0 , 0) and solve for c 1 and c 2 . This yields two equations c 1 + 2 c 2 = 0 and 2 c 1 + c 2 = 0, and it follows that c 1 = c 2 = 0 is the only solution. Hence the two vectors are linearly independent. (b) (2 , - 2) , ( - 2 , 2) Solution. Notice that ( - 2 , 2) = - (2 , - 2). Since one vector is a scalar multiple of the other, they are linearly dependent. (c) (1 , 2 , 1) , (1 , 3 , 1) , (0 , - 1 , 0) Solution. We write c 1 (1 , 2 , 1) + c 2 (1 , 3 , 1) + c 3 (0 , - 1 , 0) = (0 , 0 , 0) which yields only 2 equations c 1 + c 2 = 0 and 2 c 1 + 3 c 2 - c 3 = 0 in 3 unknowns. Hence there must be nonzero solutions: for instance, c 1 = 1 , c 2 = - 1 , c 3 = - 1 is such a solu- tion. This means the vectors are linearly dependent. (d) (1 , 0 , 0) , (1 , 1 , 0) , (1 , 1 , 1) Solution. We write c 1 (1 , 0 , 0) + c 2 (1 , 1 , 0) + c 3 (1 , 1 , 1) = (0 , 0 , 0), which yields 3 equations c 1 + c 2 + c 3 = 0 , c 1 + c 2 = 0 and c 3 = 0. It follows that c 1 = c 2 = c 3 = 0 is the only solution, and thus the vectors are linearly independent. (e) x + 1 , x 2 + 2 x, x 2 - 2 in the vector space P 2 of polynomials of degree less than or equal to 2. Solution. We write a ( x + 1) + b ( x 2 + 2 x ) + c ( x 2 - 2) = 0, and compare the coefficients of each different power of x on each side of the equation. This yields 3 equations (the first comes from looking at the constant terms, the second from the x -terms and the third from the x 2 -terms): a - 2 c = 0, a +2 b = 0 and b + c = 0. Solving for a, b, c , we see that there is a free variable, and one nonzero solution is given by a = 2 , b = - 1 , c = 1. Hence the vectors are linearly dependent. 1

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2. For each part of Problem 1, find a basis for the span of the listed vectors. What is the dimension of the span in each case? Justify your answers. Solution. (a) & (d): Since the given vectors were already linearly independent, they form a basis for their span. For the rest, we must find the largest subset of the given vectors that is linearly independent. (b): Since the two vectors (2 , - 2) and ( - 2 , 2) point in opposite directions, their span will consist of all scalar multiples of either one of them. Hence we can pick (2 , - 2) as our basis vector (any set of 1 nonzero vector is automatically linearly independent). (c): As shown above (1 , 2 , 1) = (1 , 3 , 1) + (0 , - 1 , 0), so (1 , 2 , 1) belongs to the span of (1 , 3 , 1) and (0 , - 1 , 0). Thus the span of all 3 vectors will be the same as the span of just the last 2. Clearly (1 , 3 , 1) and (0 , - 1 , 0) are linearly independent since neither is a scalar multiple of the other. Thus { (1 , 3 , 1) , (0 , - 1 , 0) } is a basis for the span.
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