Vjeko2s

# Vjeko2s - Math 33A, Midterm 2 solutions 1. We have to solve...

This preview shows pages 1–2. 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: Math 33A, Midterm 2 solutions 1. We have to solve the linear system of 1 equation in 3 variables. Its augmented coefficient matrix is ( 2- 1 1 | ) , and after dividing by 2 we get its RREF ( 1- 1 2 1 2 | ) . Alternatively, we can say that we are finding the kernel of the matrix ( 2- 1 1 ) . The solution is x 1 x 2 x 3 = s 1 2 1 + t - 1 2 1 where s,t are arbitrary parameters, i.e. the subspace (the plane) can be written as span { 1 2 1 , - 1 2 1 } . The above two vectors are linearly independent (which is always the case whenever we compute the kernel from the RREF), and thus 1 2 1 , - 1 2 1 is a basis for the given subspace. 2. Denote the matrices by A and B respectively. image( A ) = span { 1 1 1 , 2 1 , - 2- 1 } = span { 1 1 1 , 2 1 } , because - 2- 1 =- 2 1 1 1 + 2 1 . Also image( B ) = span { 3 1 2 ,...
View Full Document

## This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

### Page1 / 4

Vjeko2s - Math 33A, Midterm 2 solutions 1. We have to solve...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online