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Unformatted text preview: Solutions of Selected Problems in HW6 3.4.8 (a) For arbitrary ( a,b,c ) T R 3 , set C 1 1 1 1 + C 2 3 1 4 = a b c . The augmented matrix is 1 3 a 1 1 b 1 4 c of which RREF is 1 0 4 a 3 c 0 1 a + c 0 0 5 a + b + 4 c . There will be no solutions for C 1 ,C 2 if 5 a + b +4 c 6 = 0 . For example, ( a,b,c ) T = (0 , 1 , 1) T cannot be expressed as a linear combination of (1 , 1 , 1) T , (3 , 1 , 4) T . Thus R 3 cannot be spanned by (1 , 1 , 1) T , (3 , 1 , 4) T . (b) RREF( X ) should have no free variables so that the corresponding system for C 1 ,C 2 ,C 3 is consistent. Equivalently, you can also show det( X ) 6 = 0 . (Recall the theorem that for v 1 , , v n R n , linear independence of the vectors are equivalent to their spanning R n . Thus it suffices to show either linear independence or spanning ability to show the vectors are a basis.) (c) From the part (a), you can set...
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This note was uploaded on 07/17/2008 for the course MATH 571 taught by Professor Kim during the Spring '08 term at Ohio State.
 Spring '08
 KIM
 Linear Algebra, Algebra

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