How to find a matrix with given eigen values Find a $3\times 3$ matrix $B$ which contains $A$ as a sub-matrix and has eigen values $0$, $1$ only
where $$A=
\begin{bmatrix} -2& -12 \\1&5 \end{bmatrix}$$
I cant find a way how to construct $B$ 
I took the eigen values to be $0$, $0$, $1$ and then took $$B =
\begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&e\end{bmatrix}$$
$\implies 3+e=1\implies e=-2$
But how to take $a$, $b$, $c$, $d$?
Please help.

Let's start with your $$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&e\end{bmatrix}$$
Let's take the eigenvalues $0,1,1$. Then $\operatorname{Tr}(B)=3+e=2 \Rightarrow e=-1$. And the characteristic polynomial must be $\lambda(\lambda -1)^2=\lambda^3-2\lambda^2+\lambda$.
$$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&-1\end{bmatrix}$$
It's characteristical polynomial is:
$$\lambda^3-2\lambda^2-(ac+bd+1)\lambda - (-5ac+ad-12bc+2bd-2)$$
So:
$$\begin{cases}ac+bd+1=-1 \\ -5ac+ad-12bc+2bd-2=0\end{cases} \Rightarrow
\begin{cases}ac+bd=-2 \\ -5ac+ad-12bc+2bd=2\end{cases}$$
Let's try $c=0$ to get rid of the largest coefficients.
$$\begin{cases}bd=-2 \\ ad+2bd=2\end{cases}\Rightarrow
\begin{cases}bd=-2 \\ ad=6\end{cases}$$
Now pick for instance $a=3,b=-1, d=2$ to find:
$$B = \begin{bmatrix} -2& -12& 3\\1&5 &-1\\0&2&-1\end{bmatrix}$$
