# 1.4 - Linear Algebra -115 Solutions to Second Homework...

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Solutions to Second Homework Problem 3( b ) (Section 1 . 4)Assume that the ﬁrst vector is a linear combi- nation of the other two. Then, (1 , 2 , - 3) = a ( - 3 , 2 , 1) + b (2 , - 1 , - 1) . Doing some algebra we get that (1 , 2 , - 3) = ( - 3 a + 2 b, 2 a - b, 1 a - b ) . So, we have the system - 3 a + 2 b = 1 2 a - b = 2 1 a - b = - 3 , with corresponding matrix: - 3 2 1 2 - 1 2 1 - 1 - 3 . Some computations here show that: - 3 2 1 2 - 1 2 1 - 1 - 3 1 - 1 - 3 2 - 1 2 - 3 2 1 1 - 1 - 3 0 1 8 0 - 1 - 8 1 - 1 - 3 0 1 8 0 0 0 , i.e. the system of equations has a solution (can you ﬁnd it?), which implies that (1 , 2 , - 3) is a linear combination of ( - 3 , 2 , 1) , (2 , - 1 , - 1) . ± Problem 4( f ) (Section 1 . 4)We work as above. Say that 6 x 3 - 3 x 2 + x + 2 = a ( x 3 - x 2 + 2 x + 3) + b (2 x 3 - 3 x + 1) . Doing, again, some algebra, (and after some work) we get the system a + 2 b = 6 - a = - 3 2 a - 3 b = 1 3 a + b = 2 1

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## This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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1.4 - Linear Algebra -115 Solutions to Second Homework...

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