Mid2Sol_2

Mid2Sol_2 - and also 1 3 1 1 = 1 1 1 1 , 2 2 0 1 2 2 3 1 0...

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and also 1 1 1 = 1 2 3 1 2 1 2 1 1 0 , 2 0 1 = 1 2 3 1 2 + 1 2 1 1 0 so that 1 1 1 , 2 0 1 and 3 1 2 , 1 1 0 span the same subspace. Therefore we conclude that 0 4 k must be a linear combination of 1 1 1 and 2 0 1 . We solve the linear system 12 | 0 10 |− 4 11 | k using Gauss-Jordan elimination to obtain the RREF 2 01 | 2 00 k 2 Now we see that the system has a unique solution precisely when k 2 = 0, i.e. k = 2. 3. The vectors 1 2 1 2 0 , 0 1 2 1 2 are linearly independent because they are not scalar multiples
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This note was uploaded on 04/12/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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