2001Exam1Solutions

2001Exam1Solutions - EXAM I SOLUTIONS Math 51, Spring 2001....

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Unformatted text preview: EXAM I SOLUTIONS Math 51, Spring 2001. You have 2 hours. No notes, no books. YOU MUST SHOW ALL WORK TO RECEIVE CREDIT Good luck! Name ID number 1. (/20 points) 2. (/20 points) 3. (/20 points) 4. (/20 points) 5. (/20 points) Bonus (/10 points) Total (/100 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Kuan Ju Liu (2 and 6) Robert Sussland (3 and 7) Hunter Tart (4 and 8) Alex Meadows (10) Dana Rowland (11) Circle your section meeting time: 11:00am 1:15pm 7pm 1 1. (a) Define what it means for a set of vectors {-→ v 1 ,...,-→ v k } to be “linearly dependent”. Solution: A collection of vectors {-→ v 1 ,...,-→ v k } is linearly dependent if any one of them can be expressed as a linear combination of the others. or A collection of vectors {-→ v 1 ,...,-→ v k } is linearly dependent if there exist constants c 1 ,...c k , not all of which are zero, such that c 1-→ v 1 + ... + c k-→ v k = 0 (b) Find a linear dependence of the three vectors below, or prove that they are indepen- dent. 1 2 4 , 2 1 , - 2 5 7 Solution: We need to solve x 1 1 2 4 + x 2 2 1 + x 3 - 2 5 7 = The augmented matrix that corresponds to this system is 1 0- 2 2 2 5 4 1 7 which row reduces to 1 0 0 0 1 0 0 0 1 There are no free variables, and we see that the only solution is the trivial solution. So, the three given vectors are independent. 2 (c) Prove that three nonzero vectors in R 2 must be dependent. Solution: Solving a-→ v 1 + b-→ v 2 + c-→ v 3 =-→ 0 gives rise to a 2x3 matrix. Every 2x3 matrix has at most two pivots since there is at most one pivot per row; so there must be a column with no pivot, and thus there is a free variable. Choosing this free variable to be something other than zero and solving, we arrive at a nontrivial linear combination of the three vectors which is zero. or If the first two vectors are dependent, then we are done; if they are independent, then their span is all of R 2 . In this case, the third vector must be a linear combination of the first two, and again we conclude that the vectors are dependent....
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This note was uploaded on 08/19/2008 for the course MATH 51A taught by Professor Staff during the Spring '06 term at Stanford.

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2001Exam1Solutions - EXAM I SOLUTIONS Math 51, Spring 2001....

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