C3091424 - I , by Problem 23, I is row equivalent to B . So...

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

View Full Document Right Arrow Icon
24. (Statement of the problem.) Proof. a) Since A is row equivalent to B , by the definition, there are elementary matrices E 1 , ··· ,E k such that A = E k ··· E 1 B. (1) Since B is row equivalent to C , by the definition, there are elementary matrices F 1 , ··· ,F j ( Note: Use different notation! ) such that B = F j ··· F 1 C. (2) Substitute (2) into (1) we get A = E k ··· E 1 F j ··· F 1 C. Since E 1 , ··· ,E k ,F 1 , ··· ,F j are elementary matrices, by the definition, A is row equivalent to C . b) Suppose both A and B are nonsinngular n × n matrices. By Theorem 1.4.2, A is row equivalent to I and B is row equivalent to I . Since B is row equivalent to
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: I , by Problem 23, I is row equivalent to B . So we have that A is row equivalent to I and I is row equivalent to B . By Part (a), we get that A is row equivalent to B . / A short version of Part (b) is like the following: Suppose both A and B are nonsinngular n n matrices. By Theorem 1.4.2, A is row equivalent to I and B is row equivalent to I . By Problem 23, I is row equivalent to B . By Part (a), A is row equivalent to B . / 1...
View Full Document

Ask a homework question - tutors are online