03spr-m2sols

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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full DocumentRight Arrow Icon
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 ±rst 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
Background image of page 2
(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 su±ce for this problem. For example, any row vector will do: -→ r 1 = 3 - 1 1 3
Background image of page 3

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

View Full DocumentRight Arrow Icon
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.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/12/2010 for the course MATH 51 at Stanford.

Page1 / 13

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

This preview shows document pages 1 - 5. Sign up to view the full document.

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