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03spr-m2sols - EXAM II SOLUTIONS Math 51 Spring 2003 You...

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EXAM II SOLUTIONS Math 51, Spring 2003. You have 2 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/30 points) 2. (/30 points) 3. (/30 points) 4. (/30 points) 5. (/30 points) Bonus (/15 points) Total (/150 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: Byoung-du Kim (2 and 6) Ted Hwa (3 and 7) Jacob Shapiro (4 and 8) Ryan Vinroot (A02) Michel Grueneberg (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1
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1. (a) Find bases for the kernel and the image of the linear transformation given by T x y z = 3 x - y + z y + 2 z 3 y + 6 z Solution: Of course the kernel of the linear transformation is just the null space of the corresponding matrix; and similarly, the image is the column space. So we will solve this problem in terms of matrices. The matrix in question has columns which are the images of the standard basis vectors: T ( e 1 ) = 3 0 0 T ( e 2 ) = - 1 1 3 T ( e 3 ) = 1 2 6 So the matrix is A = 3 - 1 1 0 1 2 0 3 6 This we can row reduce to its reduced row echelon form rref( A ) = 1 0 1 0 1 2 0 0 0 The solutions to the homogeneous system are x y z = - z - 2 z z = z - 1 - 2 1 and so a basis for the null space (and thus the kernel of the linear transformation) is - 1 - 2 1 And since there are pivots in the first and second columns of rref( A ), a basis for the column space (and thus the image of the linear transformation) is 3 0 0 , - 1 1 3 2
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(b) Find any nonzero vector -→ x with the property that -→ x is perpendicular to every vec- tor in the kernel of T ; and explain how you know that the vector you supply has this property. Solution: Vectors in the row space are always perpendicular to vectors in the null space, which of course is equal to the kernel of the linear transformation. So, any vector in the row space will suffice for this problem. For example, any row vector will do: -→ r 1 = 3 - 1 1 3
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2. (a) Suppose that A is a 4x3 matrix, and that C ( A ) has dimension 2. What is the di- mension of N ( A )? Solution: The matrix A is a function from R 3 to R 4 . So, the Rank-Nullity theorem tells us that dim C ( A ) + dim N ( A ) = 3 Since the dimension of C ( A ) is given to be 2, we know that N ( A ) must have dimen- sion 1. (b) Suppose that N ( M 2 ) = { -→ 0 } . Show that N ( M 2 M 1 ) = N ( M 1 ). Solution: We need to show that M 2 M 1 -→ x = -→ 0 ⇐⇒ M 1 -→ x = -→ 0 ( ) If M 1 -→ x = -→ 0 , then M 2 M 1 -→ x = M 2 ( M 1 -→ x ) = M 2 -→ 0 = -→ 0 .
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