<s>[INST] Finding $B,C$ such that $B\left[\begin{smallmatrix}1&2\\4&8\end{smallmatrix}\right]C=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ 
State $B,C$, such that $B\begin{bmatrix} 1 & 2  \\ 4 & 8    \end{bmatrix}C=\begin{bmatrix}
1 & 0  \\
0 & 0  
\end{bmatrix}$

I tried some things, but I ended up with non invertible matrices, so I stopped for the moment.
Is there a quick way to guesswork here? (it's an assignment so, maybe some "nice" values for the entries will do)
I don't even know how I multiply a matrix to get 0-Entries with invertible matrices.
So maybe let's start at that part?
 [/INST] $B,C$ are not unique.
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = B^{-1} 
\begin {bmatrix} 1&0\\0&0 \end{bmatrix}C^{-1}$
Suppose 
$B^{-1} = \begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix}$ and
$C^{-1} = \begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix}$ 
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = $$\begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} 1&0\\0&0 \end{bmatrix}\begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22}\end{bmatrix}\\
\begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} c_{11}&c_{12}\\0&0 \end{bmatrix}
\\
\begin {bmatrix} b_{11}c_{11}&b_{11}c_{12}\\b_{21}c_{11}&b_{21}c_{12} \end{bmatrix}$
$b_{11}c_{11} = 1\\
b_{11}c_{12}= 2\\
c_{12} = 2c_{11}\\
b_{21}c_{11} = 4b_{11}\\
$
Choose the other elements to make your calculation easy.  i.e. $\det B^{-1} = 1$
$B = \begin {bmatrix} 1&0\\-4&1 \end{bmatrix}$
$C = \begin {bmatrix} 1&-2\\0&1 \end{bmatrix}$
Should work.
Completely different approach.
Daigonlize $\begin{bmatrix} 1&2\\4&8 \end{bmatrix}$
$P^{-1}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}P = \begin{bmatrix} 9&0\\0&0 \end{bmatrix}$
$\frac 1{81}\begin{bmatrix} 1&2\\-4&1\end{bmatrix}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}\begin{bmatrix} 1&-2\\4&1\end{bmatrix} = \begin{bmatrix} 1&0\\0&0 \end{bmatrix}$
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