hw9solutions

hw9solutions - Math 215 HW #9 Solutions 1. Problem 4.4.12....

Info iconThis preview shows pages 1–4. 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

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: Math 215 HW #9 Solutions 1. Problem 4.4.12. If A is a 5 by 5 matrix with all | a ij | 1, then det A . Volumes or the big formula or pivots should give some upper bound on the determinant. Answer: Let ~v i be the i th column of A . Then | ~v i | = q a 2 1 i + a 2 2 i + a 2 3 i + a 2 4 i + a 2 5 i . Since | a ij | 1 for each i and j , its also true that each a 2 ij 1. Hence, the right hand side of the above equation is no bigger than 1 + 1 + 1 + 1 + 1 = 5, and so we see that | ~v i | 5 . This means that each edge of the 5-dimensional box spanned by the columns of A is no longer than 5, meaning that the volume of that box can be no bigger than 5 5 = 5 5 / 2 = 25 5 . Since | det A | is exactly equal to the volume of the box spanned by the columns of A , this reasoning implies that | det A | 25 5 55 . 9 . Another way to get an upper bound is to use the big formula for the determinant from p. 212. Notice that, for any 5 by 5 matrix, there are 5! = 120 terms in the sum. When all entries of A are smaller than 1 (in absolute value), it must be the case that each of the 120 terms in the sum is smaller than 1 (in absolute value). Therefore, the whole sum (which is just the determinant), must be no bigger than 120 in absolute value. 2. Problem 4.4.18. Find A- 1 from the cofactor formula C T / det A . Use symmetry in part (b): (a) A = 1 2 0 0 3 0 0 4 1 . (b) A = 2- 1- 1 2- 1- 1 2 . (a) We need to determine the various cofactors of A to find the cofactor matrix C : C 11 = (- 1) 1+1 det A 11 = 3 0 4 1 = 3 C 12 = (- 1) 1+2 det A 12 =- 0 0 0 1 = 0 C 13 = (- 1) 1+3 det A 13 = 0 3 0 4 = 0 C 21 = (- 1) 2+1 det A 21 =- 2 0 4 1 =- 2 1 C 22 = (- 1) 2+2 det A 22 = 1 0 0 1 = 1 C 23 = (- 1) 2+3 det A 23 =- 1 2 0 4 =- 4 C 31 = (- 1) 3+1 det A 31 = 2 0 3 0 = 0 C 32 = (- 1) 3+2 det A 32 =- 1 0 0 0 = 0 C 33 = (- 1) 3+3 det A 33 = 1 2 0 3 = 3 Therefore, C = C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 = 3- 2 1- 4 3 . Also, since weve already computed the relevant cofactors, its easy to find det A : det A = a 11 C 11 + a 12 C 12 + a 13 C 13 = 1 3 + 2 0 + 0 0 = 3 . Therefore, A- 1 = 1 det A C T = 1 3 3- 2 0 1- 4 3 = 1- 2 / 3 0 1 / 3- 4 / 3 1 (b) Again, we need to determine the various cofactors of A to find the matrix C , but now we can use the fact that A is symmetric, which implies that A ij = A T ji and so C ij = (- 1) i + j det A ij = (- 1) i + j det A T ji = (- 1) j + i det A ji = C ji . Thus, we only need to compute the C ij for which i j : C 11 = (- 1) 1+1 det A 11 = 2- 1- 1 2 = 3 C 12 = (- 1) 1+2 det A 12 =-- 1- 1 2 = 2 C 13 = (- 1) 1+3 det A 13 =- 1 2- 1 = 1 C 22 = (- 1) 2+2 det A 22 = 2 0 0 2 = 4 C 23 = (- 1) 2+3 det A 23 =- 2- 1- 1 = 2 C 33 = (- 1) 3+3 det A 33 = 2- 1- 1 2 = 3 2 Therefore, C = C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 = 3 2 1 2 4 2 1 2 3 ....
View Full Document

This document was uploaded on 08/20/2011.

Page1 / 10

hw9solutions - Math 215 HW #9 Solutions 1. Problem 4.4.12....

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