<s>[INST] $3\times 3$ matrix with eigenvalues are given 
If $\displaystyle P=\begin{bmatrix}
0 & -2 & 3\\ 
 -1&  1& -1\\ 
a & 2 & b
\end{bmatrix}$ for some $a,b\in \mathbb{R},$ suppose $1$ and $2$ are eigen values of $P$ 
and $\displaystyle P\begin{bmatrix}
 1\\ 
 0\\ 
-3
\end{bmatrix} =\begin{bmatrix}
 3\\ 
 0\\ 
-3
\end{bmatrix},$ Then $\displaystyle P^4\begin{bmatrix}
 1\\ 
 -1\\ 
0
\end{bmatrix} =$

$\bf{My\; Try::}$ Here $$\displaystyle \begin{bmatrix}
0 & -2 & 3\\ 
 -1&  1& -1\\ 
a & 2 & b
\end{bmatrix}\begin{bmatrix}
 1\\ 
 0\\ 
-3
\end{bmatrix} =\begin{bmatrix}
 3\\ 
 0\\ 
-3
\end{bmatrix}$$
After multiplication we get $$\begin{bmatrix}
 3\\ 
 1\\ 
a-b
\end{bmatrix}=\begin{bmatrix}
 3\\ 
 1\\ 
-3
\end{bmatrix}$$
So we get $a-3b=-3$
Now how can i solve it after that, Help required, Thanks
 [/INST] $$tr(P)=\lambda_1+\lambda_2+\lambda_3$$
$$b+1=1+2+\lambda_3\implies\lambda_3=b-2$$
$$det(P)=\lambda_1\lambda_2\lambda_3$$
$$-a-2b-6=2(b-2)\implies a+4b=-2$$
on the other hand 
$$a-3b=-3$$
if 
$$\displaystyle \begin{bmatrix}
0 & -2 & 3\\ 
 -1&  1& -1\\ 
a & 2 & b
\end{bmatrix}\begin{bmatrix}
 1\\ 
 0\\ 
-3
\end{bmatrix} =\begin{bmatrix}
 -9\\ 
 2\\ 
-3
\end{bmatrix}$$
</s>