hw3 - R 3 are actually a basis. Hence, we now have a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Winter 2011 Math 67 Linear Algebra Homework 3 Problems 2.3, 2.4, 2.8, 2.13 and 2.14 from Axler. Problem A. Recall that if we have three vectors ( x 1 ,y 1 ,z 1 ), ( x 2 ,y 2 ,z 2 ), ( x 3 ,y 3 ,z 3 ) in R 3 then the volume of the parallelapiped they generate is given by the absolute value of the following expression x 1 ± ± ± ± y 2 z 2 y 3 z 3 ± ± ± ± - y 1 ± ± ± ± x 2 z 2 x 3 z 3 ± ± ± ± + z 1 ± ± ± ± x 2 y 2 x 3 y 3 ± ± ± ± where ± ± ± ± a b c d ± ± ± ± = ad - bc . We deemed it clear that the three vectors above span R 3 if and only if the volume of the parallelapiped they give rise to has positive volume. By a result in class we know that three vectors that span
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: R 3 are actually a basis. Hence, we now have a convienient test for a collection of three vectors to be a basis. Using the procedure above, determine if the following collections of three vectors form a basis for R 3 : ( a ) (1 , 1 , 0) (1 , , 1) (1 , 3 ,-2) ( b ) (1 , 1 , 0) (0 , 1 , 1) (3 , , 1) ( c ) (1 , 2 , 3) (4 , 5 , 6) (7 , 8 , 9) 1...
View Full Document

This note was uploaded on 11/13/2011 for the course MATH 67 taught by Professor Schilling during the Winter '07 term at UC Davis.

Ask a homework question - tutors are online