Vector equation with no solution There exists a real number $k$ such that the equation
$\begin{pmatrix} -1 \\ -2 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} + s\begin{pmatrix} -4 \\ k \end{pmatrix}$
does not have any solutions in $t$ and $s$. Find $k$.
I am confused as to how I will find k so this equation has NO solutions. Any help is appreciated. 
Thanks
 A: In order for the system of equations to have no solutions, the constituent equations must be linearly dependent.  
This occurs when $3k=(-2)(-4)\implies k=\frac{8}{3}$.
$$\bbox[5px,border:2px solid #C0A000]{\text{The system has no solutions when}\,\, k=8/3}$$
A: Given $$\begin{bmatrix}-1 \\ -2\end{bmatrix} + t\begin{bmatrix}3\\-2\end{bmatrix} = \begin{bmatrix}5 \\ 0\end{bmatrix} + s\begin{bmatrix}-4\\k\end{bmatrix}$$ implies
$$-s\begin{bmatrix}-4\\k\end{bmatrix} + t\begin{bmatrix}3 \\- 2\end{bmatrix} = \begin{bmatrix}5\\0\end{bmatrix} - \begin{bmatrix}-1\\-2\end{bmatrix}$$ which is equivalent to $$s\begin{bmatrix}4\\-k\end{bmatrix} + t\begin{bmatrix}3 \\- 2\end{bmatrix} = \begin{bmatrix}5\\0\end{bmatrix} + \begin{bmatrix}1\\2\end{bmatrix}$$ or rather $$\begin{bmatrix}4s + 3t\\-ks - 2t\end{bmatrix} = \begin{bmatrix}6\\2\end{bmatrix}.$$ In matrix form, this system is equivalent to $$\begin{bmatrix}4 & 3 & 6\\-k & -2 & 2\end{bmatrix}$$ Applying row reduction to this system allows us to get the reduced row echelon form $$\text{rref}\begin{bmatrix}4 & 3 & 6\\-k & -2 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 & {3 \over 2} - {3(3k+4) \over 2(3k-8)}\\0 & 1 & {2(3k+4) \over 3k-8}\end{bmatrix}.$$
The system should have a solution for all $k \in \mathbb{R}\backslash \{{8 \over 3}\}$ because $k = {8 \over 3}$ makes the system undefined. So $k = {8 \over 3}$ makes the system have no solution.
