$A=\begin{pmatrix}a&c&-1\\ 1-c&-a&0\\ 5&3&b\end{pmatrix}$ has an eigenvector $(-1,1,-1)'$, and $|A|=0$, can we find $a,b,c$? 
$A=\begin{pmatrix}a&c&-1\\ 1-c&-a&0\\ 5&3&b\end{pmatrix}$ has an eigenvector $(-1,1,-1)'$, and $|A|=0$, can we find $a,b,c$?

By $Ax=tx$ for some $t$, we see $c=1+t-a, b=t-2$ (with the other equality superfluous), and then
$0=|A|=t^3-(2a+1)t^2+(3a+1)t-6a$. Oh my god, we can not do anymore.

Since $\det(A)=0$, then we have
$$-ba^{2}-5a+(c-1)(bc+3)=0,\quad (1)$$
Since $(-1,1-1)$ is an eigenvector for some eigenvalue $k$ of $A$ we have
$$-a+c+k=-1,\quad (2)$$
$$-a+c-k=1,\quad (3)$$
$$-b+k=2,\quad (4)$$
Solving the non-linear system with equations $(1),(2),(3)$ and $(4)$ with Mathematica we have one solution
$$(a,b,c,k)=(3,-3,3,-1).$$
It works, since

*

*$\det(A)=\det\begin{pmatrix}3&3&-1\\ 1-3&-3&0\\ 5&3&-3\end{pmatrix}=0$.

*$Av=\begin{pmatrix}3&3&-1\\ 1-3&-3&0\\ 5&3&-3\end{pmatrix}\begin{pmatrix}-1\\1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}=-1\begin{pmatrix} -1\\1\\-1\end{pmatrix}=kv$.

So a priori with that information we can find the values $a,b,c$. Of course, it would be interesting to be able to avoid using the software.
