If I were given a matrix $A$, $$ A = \begin{bmatrix} 5 & -2 & 3 \\ 0 & y & 0 \\ x & 7 & z \\ \end{bmatrix} $$ with corresponding eigenvalues $1, 7,$ and $−4$, respectively, find $x, y,$ and $z$.
I am familiar with finding eigenvalues and eigenvectors for matrices, but I'm not sure how to find $x, y, z$ for the matrix itself from the given eigenvalues. I'm currently stuck on using the determinant formula for each eigenvalues. $$\begin{split} 0&=\det(A−I)\\ 0&=\det(A−7I)\\ 0&=\det(A+4I)\\ \end{split}$$
$\det(A-\lambda I)=(5-\lambda)(y-\lambda)(z-\lambda)-3x(y-\lambda)=(y-\lambda)(\lambda^2-(5+z)\lambda+5z-3x)$. This polynomial in $\lambda$ has roots $1, 7, -4$.
The root of the linear term is $y$. If $y=-4$, the remaining roots are $1, 7$, which would mean $5+z=8$ and $5z-3x=7$, so $z=3, x=\frac 83$.
If $y=1$, the remaining roots are $7, -4$, which would mean $5+z=3$ and $5z-3x=-28$, so $z=-2, x=6$.
If $y=7$, the remaining roots are $1, -4$, which would mean $5+z=-3$ and $5z-3x=-4$, so $z=-8, x=-12$.
Confirm (by plugging them into the matrix and taking the determinant of $A-\lambda I$) that all three of these values for $(x, y, z)$ give you the required eigenvalues and you're done.