How to find specific variables that cause vectors to be linearly independent / dependent The vectors
$v= \begin{bmatrix}
5\\
2\\
7\\
\end{bmatrix}, u = \begin{bmatrix}
4\\
4\\
13+k\\
\end{bmatrix}, \text{and } w = \begin{bmatrix}
-4\\
-2\\
-6\\
\end{bmatrix} $
are linearly independent if and only if $k \neq$ what?
How exactly can I go about solving this?
 A: Hint:
If you don't want to use the determinant you can write
$$l_1{\bf v} + l_2{\bf u} + l_3{\bf w} + l_4{\bf [0,0,1]^T = 0}, (\text{where   } l_4/l_2 = k)$$
This will be an overdetermined equation system and it will have a least-squares solution:
$$\min_{\forall l_k} \|l_1{\bf v} + l_2{\bf u} + l_3{\bf w} + l_4{\bf [0,0,1]^T\|_2 }$$
when $l_4/l_2 = k$ gives a perfect fit.
A: Hint:
If you have three vectors in three dimensions, then they are linearly dependent if and only if the determinant, of the 3-by-3 matrix whose columns (or rows) are your vectors, is zero.
(The determinant gives the volume of the parallelepiped spanned by its columns (or rows)).
In general, if you have a set of $n$ vectors in $n$-dimensional space, they are linearly dependent if and only if the determinant, of the $n$-by-$n$ matrix whose columns (or rows) are your $n$ vectors, is zero.
A: As Fly by Night mentioned, these vectors will be linearly dependent if and only if $$\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} = 0.$$
\begin{align*}
\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} &= 5[(4)(-6) - (13+k)(-2)] - 4[(2)(-6) - (7)(-2)] + (-4)[(2)(13+k)-(7)(4)] \\
&=2k+10
\end{align*}
Therefore the vectors will be linearly dependent if and only if $$ 2k+10 = 0.$$
