Midterm 2 ReviewSolutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 donejust 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) 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) , (1 , 1 , 0) , (1 , 1 , 1) Solution. We write c 1 (1 , , 0) + c 2 (1 , 1 , 0) + c 3 (1 , 1 , 1) = (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 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....
View Full Document

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online