exam1-solution

# exam1-solution - Exam 1 — Version 1 – Solutions Math...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exam 1 — Version 1 – Solutions Math 340L, Spring 2007, Purcell 1. (5 points) Let A be the matrix: 1 3 6 2 6 12 0 4 1 Find properties of b = b 1 b 2 b 3 such that b is in the span of the columns of A . Show all work. Solution: We row reduce the augmented matrix: 1 3 6 b 1 2 6 12 b 2 0 4 1 b 3 Add- 2 times the first row to the second to cancel the 2 in the first column, second row. 1 3 6 b 1 0 0 0- 2 b 1 + b 2 0 4 1 b 3 Swap the second and third rows. 1 3 6 b 1 0 4 1 b 3 0 0 0- 2 b 1 + b 2 Note there is a pivot in the first and second columns. The vector b will be in the span of the columns of A if and only if there is no pivot in the last column, or iff- 2 b 1 + b 2 = 0 . 1 2. The problems below concern the matrix: A = bracketleftbigg 1 2 6 0 2 6 bracketrightbigg (a) (3 points) Describe all solutions to A x = in parametric vector form. Show your work. Solution: We row reduce A . A = bracketleftbigg 1 2 6 0 2 6 bracketrightbigg ∼ bracketleftbigg 1 0 0 0 2 6 bracketrightbigg ∼ bracketleftbigg 1 0 0 0 1 3 bracketrightbigg Now the rows of the reduced matrix give the equations: x 1 = 0 , x 2 + 3 x 3 = 0 , and x 3 is free. In parametric vector form: x 1 x 2 x 3 = x 3 - 3 1 (b) (2 points) Does A x = b have at least one solution for every possible b ? Why or?...
View Full Document

## This note was uploaded on 03/20/2008 for the course M 340L taught by Professor Pavlovic during the Spring '08 term at University of Texas.

### Page1 / 8

exam1-solution - Exam 1 — Version 1 – Solutions Math...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online