Exercise_6

# Exercise_6 -  , b be   3 3 3   Then a direct...

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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310 Linear Algebra and Applications (Fall 2007) Solution of Exercise 6 Name: Student ID: Class: Determine which of the linear systems have a solution by using Theorem 0.6 (cf. Lecture note 11, Page 70) 1 2 1 1 1 0 2 1 - 1 x 1 x 2 x 3 = 3 3 3 Solution: Let A be 1 2 1 1 1 0 2 1 - 1
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Unformatted text preview:  , b be   3 3 3   Then a direct computation shows that A is row equivalent to   0 1 1 1 1 0 0 0 0   and [A | b] is row equivalent to   0 1 1 1 1 0 3 0 0 0-3   so rank(A) is 2 and rank([A | b]) is 3. Therefore by theorem 0.6, the given system has no solution....
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## This note was uploaded on 05/18/2010 for the course MATHEMATIC MAT2310 taught by Professor Dr.jeffchak-fuwong during the Spring '06 term at CUHK.

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