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

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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
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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...
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