Quiz3-Solutions

Quiz3-Solutions - 2-2 4 4-6 7 -2 2 h The vectors are...

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Name:. ......................................................................................... Math 331 - Quiz 3 - Solutions Thursday, September 22 1. Describe all of the solutions to A x = 0 in parametric form A = 1 - 4 - 2 0 3 - 5 0 0 1 0 0 - 1 0 0 0 0 1 - 4 0 0 0 0 0 0 The free variables are x 2 ,x 4 , and x 6 . The matrix is in echelon form but not reduced row echelon form. From the echelon form we can use the “back substitution” method to express the basic variables in terms of the free variables. The equations are x 1 - 2 x 3 + 3 x 5 = 4 x 2 + 5 x 6 x 3 = x 6 x 5 = 4 x 6 Subbing the last two equations into the first gives us the basic variable expressed in terms of the free variables. x 1 = 4 x 2 - 5 x 6 x 2 is free x 3 = x 6 x 4 is free x 5 = 4 x 6 x 6 is free As a vector in R 6 we express this in parametric form as x = 4 x 2 - 5 x 6 x 2 x 6 x 4 4 x 6 x 6 = x 2 4 1 0 0 0 0 + x 4 0 0 0 1 0 0 + x 6 - 5 0 1 0 4 1
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2. Find the value(s) of h for which the vectors are linearly dependent
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Unformatted text preview: 2-2 4 4-6 7 -2 2 h The vectors are linearly dependent if there is a non-trivial linear combination of them which gives the zero vector. So we want to solve. 2 4-2 0-2-6 2 4 7 h II + I---- III-2 I 2 4-2-2-1 h + 4 0 III-(1 / 2) II------- 2 4-2-2 h + 4 0 So the vectors are linearly dependent when h =-4. We can check and see that when h =-4 we have that v 1 =-v 3 so there is a dependence relation in this case. if I saw this dependence relation originally, I would know that h =-4 makes the vectors linearly dependent, but I would still want to go through the calculations above to show that no other h gives a dependent set of vectors....
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Quiz3-Solutions - 2-2 4 4-6 7 -2 2 h The vectors are...

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