winter 2007assign2-sol

winter 2007assign2-sol - MATH235 SOLUTIONS TO ASSIGNMENT 2....

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: MATH235 SOLUTIONS TO ASSIGNMENT 2. 1. Section 3.1 14) Compute the determinant of the following matrix by cofactor expansions. At each step choose a row or a column that involves the least amount of computation. 6 3 2 4 9- 4 1 8- 5 6 7 1 3 4 2 3 2 Solution. 6 3 2 4 9- 4 1 8- 5 6 7 1 3 4 2 3 2 = 1 * 6 3 2 4 9- 4 1 3 4 2 3 2 = 3 * 3 2 4- 4 1 2 3 2 = 3 * 3 *- 4 1 3 2 + 2 * 2 4- 4 1 = 3 * (3 * (- 4 * 2- 3 * 1) + 2 * (2 * 1- (- 4) * 4)) = 9 . Section 3.2 10) Find the determinant of the following matrix by row reduction to echelon form. 1 3- 1- 2 2- 4- 1- 6- 2- 6 2 3 9 3 7- 3 8- 7 3 5 5 2 7 Solution. 1 1 3- 1- 2 2- 4- 1- 6- 2- 6 2 3 9 3 7- 3 8- 7 3 5 5 2 7 = 1 3- 1- 2 2- 4- 1- 6 3 5- 2 8- 1- 4 8 2 13 R 3 + 2 * R 1 → R 3 R 4- 3 * R 1 → R 4 R 5- 3 * R 1 → R 5 = 1 3- 1- 2 2- 4- 1- 6 3 5- 4 7- 7 1 R 4 + R 2 → R 4 R 5 + 2 * R 2 → R 5 =- 1 * 1 3- 1- 2 2- 4- 1- 6- 4 7- 7 3 5 1 ( R 4 <- > R 3 ) =- 1 * 1 * 2 * (- 4) * 3 * 1 = 24 . 22) Use the determinant to find out if the following matrix is invertible. 5- 1 1- 3- 2 5 3 Solution. 5- 1 1- 3- 2 5 3 = 5 * (- 3 * 3- (- 2) * 5)- 1 * (0 * 3- (- 1) * 5) = 0 The matrix is not invertible since the determinant is zero....
View Full Document

This note was uploaded on 05/23/2010 for the course MATH 235 taught by Professor Celmin during the Winter '08 term at Waterloo.

Page1 / 6

winter 2007assign2-sol - MATH235 SOLUTIONS TO ASSIGNMENT 2....

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

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