Linear Algebra Solutions 28

Linear Algebra Solutions 28 - Also X= Y = Z= 1 1 1 (X + Y )...

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Also X = 1 2 ( X + Y ) + 1 2 ( X + Z ) - 1 2 ( Y + Z ) , Y = 1 2 ( X + Y ) - 1 2 ( X + Z ) + 1 2 ( Y + Z ) , Z = - 1 2 ( X + Y ) + 1 2 ( X + Z ) + 1 2 ( Y + Z ) , so h X, Y, Z i ⊆ h X + Y, X + Z, Y + Z i . Hence h X, Y, Z i = h X + Y, X + Z, Y + Z i . 3. Let X 1 = 1 0 1 2 , X 2 = 0 1 1 2 and X 3 = 1 1 1 3 . We have to decide if X 1 , X 2 , X 3 are linearly independent, that is if the equation xX 1 + yX 2 + zX 3 = 0 has only the trivial solution. This equation is equivalent to the folowing homogeneous system x + 0 y + z = 0 0 x + y + z = 0 x + y + z = 0 2 x + 2 y + 3 z = 0 . We reduce the coeFcient matrix to reduced row–echelon form: 1 0 1 0 1 1 1
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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