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exam1sol - Math 51 Spring 2009 Exam 1 Solutions Problem...

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Math 51 — Spring 2009 — Exam 1 Solutions Problem 1. (15 pts.) Mark the following statements as either TRUE or FALSE . If a statement is false, give a simple counterexample. If a statement is true, give a justification. Read the questions very carefully! Three points per part: 1 point for the answer and 2 points for the justification (except in parts (b), (c), (d), where these point values were swapped). a) If the reduced row echelon form of a matrix A has a pivot in every column, then the null space N ( A ) of A is empty. TRUE FALSE We know that the null space of any matrix always contains the zero vector, so in particular, it is never empty. (In fact, under the given assumption on A , then the null space of A contains only the zero vector; it does not contain any nonzero vectors.) Notes: A very common mistake was to confuse the set containing only the zero vector with the empty set . b) The cross product of two vectors in 3 has length zero if and only if the two vectors are linearly dependent. TRUE FALSE Since k v × w k = k v kk w k sin θ , we see that if v × w has length 0, then either v or w is the zero vector, or sin θ = 0. If v = 0 or w = 0 , then { v , w } is linearly dependent since any set containing the zero vector is linearly dependent. If sin θ = 0, then this means that v and w lie on the same line, so again { v , w } is linearly dependent. On the other hand, if v , w are dependent, then one of them is a scalar multiple of the other, say v = c w . But then v × w = ( c w ) × w = c ( w × w ) = c ( 0 ) = 0 , and so v × w has length 0. Notes: Many tried to prove this true simply by giving a single example — but one example is never enough to show that a statement about arbitrary vectors is true. c) The dot product of two vectors is zero if and only if the two vectors are orthogonal. TRUE FALSE This is the definition of orthogonal (see page 26 of the Levandosky text). d) For any k × n matrix A , dim N ( A ) + dim C ( A ) = n . TRUE FALSE This is the statement of the Rank-Nullity Theorem. e) For any v 1 , v 2 in n , if dim Span { v 1 , v 2 } = 0, then v 1 = v 2 . TRUE FALSE The only subspace of n of dimension 0 is { 0 } , so the given assumption tells us that Span { v 1 , v 2 } = { 0 } . But v 1 , v 2 Span { v 1 , v 2 } , so we must have v 1 = v 2 = 0 , and therefore v 1 = v 2 . 1
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Problem 2. (12 pts.) Consider the following system of equations: x + 3 y + z = b 1 2 x - z = b 2 3 x + 3 y + 2 z = b 3 where x , y and z are unknowns, and b 1 , b 2 and b 3 are real numbers. a) For what values of b 1 , b 2 and b 3 is x = 0 y = 1 z = 2 a solution to the above system? (4 points) Setting x = 0, y = 1 and z = 2 in the given system yields b 1 = 0 + 3 · 1 + 2 = 5 b 2 = 2 · 0 - 2 = - 2 b 3 = 3 · 0 + 3 · 1 + 2 · 2 = 7 b) For what values of b 1 , b 2 and b 3 is x = 0 y = 1 z = 2 the unique solution to the above system? (4 points) From part (a) we know that only ( b 1 , b 2 , b 3 ) = (5 , - 2 , 7) gives ( x, y, z ) = (0 , 1 , 2) as a solution. To check uniqueness of this solution, we find the null space of the coefficient matrix (call it A ). Row reducing gives: 1 3 1 2 0 - 1 3 3 2 ; 1 3 1 0 - 6 - 3 0 - 6 - 1 ; 1 3 1 0 1 1 2 0 1 1 6 ; 1 0 - 1 2 0 1 1 2 0 0 - 1 3 ; 1 0 0 0 1 0 0 0 1 Thus N ( A ) = { 0 }
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