Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$ 
\begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A &B \\ 
C & D
\end{bmatrix}) $ but it didn't help me.

The equation
$$\begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}
=0$$
is equivalent to $p_A(-x)=0$, where $p_A$ is the characteristic polynomial of
$$A=\begin{pmatrix}
0 & a & b & c \\ 
a & 0 & b & c \\ 
m &n  & 0 &p \\ 
 m&  n&  p& 0
\end{pmatrix};$$hence the roots of your equation must be the opposite of the eigenvalues of $A$. The condition that $a+m+n=b+c+p$ is equivalent to $a-b-c=-m-n+p$, which tells you that $(1,1,-1,-1)$ is an eigenvector, with associated eigenvalue $a-b-c$. Moreover $-a$ and $-p$ are obviously eigenvalues, and the trace of the matrix is $0$; hence the sum of the eigenvalues is zero, which means that the last eigenvalue must be $b+c-a+a+p=b+c+p$.
So the solutions to your equation are $a$, $p$, $-(b+c+p)$ and $b+c-a$.
