Another question of finding eigenvalues with parameters Let
$$A=\left(\begin{matrix}a&b&c\\a-s&b+s&c\\a-t&b&c+t\end{matrix}\right)$$
Find A's eigenvalues.
So, I did some row operations without changing the value of A's determinant, so that I got 
$$A'=\left(\begin{matrix}a&b&c\\-s&0&0\\-t&0&t\end{matrix}\right)$$
But I still can't get a simple way for finding the eigenvalues. Are there more row operations I can do? 
Thanks in advance for any assistance!  
 A: According to Maple, the characteristic polynomial of $A$ is
$$\begin{align}
x^3 &{}+(-c-a-t-b-s)x^2 \\&{}+ (-a^2+2as+st+at+sc+tc+ab+bt)x \\&{}+ (tac-asc-2ast+scb-abt+a^2t-tsc-tcb),\end{align}$$ and it does not factor. This means the problem is pretty hopeless. The constant term is also clearly not $-\det(A')$, so either you mistyped the matrix $A$ in the first place, or you made errors in the row operations leading to $A'$ (but which would not lead to solving this problem anyway if done correctly, just to get the right constant term of the characteristic polynomial).
Added. With the now modified matrix, the simplest thing is to conjugate by $I_3-E_{1,2}-E_{1,3}$ (with $E_{i,j}$ an elementary matrix; conjugations means left-multiply by this matrix and right-multiply by the inverse $I_3+E_{1,2}+E_{1,3}$); this transforms $A$ into the similar matrix
$$
  A'=\begin{pmatrix}a+b+c&b&c\\0&s&0\\0&0&t\end{pmatrix},
$$
which being triangular clearly has eigenvalues $a+b+c$, $s$, and $t$, and so has$~A$.
