2002Exam2Solutions

2002Exam2Solutions - EXAM II SOLUTIONS Math 51, Spring...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
EXAM II SOLUTIONS Math 51, Spring 2002. 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: Tarn Adams (2 and 6) Mariel Saez (3 and 7) Yevgeniy Kovchegov (4 and 8) Heaseung Kwon (A02) Alex Meadows (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) Without computing the inverse directly, and without row reducing, explain how you know that the matrix A given below is invertible. A = 3 0 2 0 2 4 0 0 1 Solution: We can compute the determinant of this matrix using minors: det A = 3det ± 2 4 0 1 - 0det ± 0 4 0 1 + 2det ± 0 2 0 0 = 6 6 = 0 Since the determinant of the matrix is not zero, we know that it must be invertible. (b) Find A - 1 . Solution: We row reduce ( A | I ): 3 0 2 0 2 4 0 0 1 1 0 0 0 1 0 0 0 1 3 0 0 0 2 0 0 0 1 1 0 - 2 0 1 - 4 0 0 1 (1 st ) - 2(3 rd ) (2 nd ) - 4(3 rd ) (3 rd ) 1 0 0 0 1 0 0 0 1 1 / 3 0 - 2 / 3 0 1 / 2 - 2 0 0 1 (1 st ) / 3 (2 nd ) / 2 (3 rd ) The end result is ( I | A - 1 ); so, A - 1 = 1 / 3 0 - 2 / 3 0 1 / 2 - 2 0 0 1 2
Background image of page 2
2. For each of the following, either (i) find a matrix (or matrices) satisfying the given de- scriptions, or (ii) prove that such a matrix (or matrices) cannot exist. (a) The column space C ( M 1 ) contains the vectors 4 3 7 4 , 1 3 7 8 and the null space N ( M 1 ) contains the vectors 1 3 2 , 4 5 - 6 Solution: The given null space vectors have three components; since the null space is a subspace of the domain, we conclude that the domain is R 3 . Also, since the two null space vectors are independent, we conclude that the null space is at least two dimensional; similarly, we conclude that the column space must be at least two dimensional. However, the Rank-Nullity theorem says that the dimension of the null space plus the dimension of the column space must equal the dimension of the domain; which, with the given data, would be impossible (since 3 < 2 + 2).
Background image of page 3

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

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

This note was uploaded on 08/19/2008 for the course MATH 51A taught by Professor Staff during the Spring '06 term at Stanford.

Page1 / 13

2002Exam2Solutions - EXAM II SOLUTIONS Math 51, Spring...

This preview shows document pages 1 - 4. Sign up to view the full document.

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