# find z such that $\vec{u}$ and $\vec{v}$ are linearly independent

The problem is given as, "Determine the values of z such that the vectors $\vec{u} = \pmatrix{-1\\z}$ and $\vec{u} = \pmatrix{z\\-1 + z}$ are linearly independent.

Here is my work...

$\pmatrix{-1 & z& \\ z& (-1 + z)& }$

$\pmatrix{-1 & z& \\ 0& (z^2+z-1)& }$

therefore, $z^2+z-1 \neq 0$

$z \neq \frac{1}{2}(-1-\sqrt{5})$

$z \neq \frac{1}{2}(\sqrt{5} + 1)$

Does that seem right?

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Looks good to me. – user7530 Feb 1 '13 at 5:13
I don't get how you get to your second parenthesis from the first. Linear independency means non-zero determinant, and you got it right. – mbaitoff Feb 1 '13 at 6:13
@mbaitoff, he multiplied first row to $z$ and added it to 2nd row. – Kaster Feb 1 '13 at 6:36
It is good except you mess up the quadratic formula slightly. The answer is $z\neq \frac{1}{2}(-1\pm\sqrt{5})$. Note the $1$ does not change signs. – user45150 Feb 1 '13 at 7:35

$$z^2 + z - 1 \ne 0 \Rightarrow \\ z^2 + z +\frac{1}{4} \ne \frac{5}{4} \Rightarrow \\ \left(z+\frac{1}{2}\right)^2 \ne \frac{5}{4} \Rightarrow\\ z \ne -\frac{1}{2} + \frac{\sqrt{5}}{2}, z \ne -\frac{1}{2} - \frac{\sqrt{5}}{2}$$