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.
 A: 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 $b+c-a$.
A: Subtract the first line from the second:
$$\begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}=\begin{vmatrix}
x-a & a-x & 0 &0 \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
m & n&  p& x
\end{vmatrix}=(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
m & n&  p& x
\end{vmatrix}=$$
$1.$ Multiply the first by $-a$ and add on the second
$2.$ Multiply the first by $-m$ and add on the third
$3.$ Multiply the first by $-m$ and add on the forth
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &m+ n&  p& x
\end{vmatrix}=$$
Forth line minus third:
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &0&  p-x& x-p
\end{vmatrix}=(x-a)(x-p)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &0&  -1& 1
\end{vmatrix}=$$
Laplace theorem on the first column
$$(x-a)(x-p)\begin{vmatrix}
x+a & b &c \\ 
m+n  & x &p \\ 
0&  -1& 1
\end{vmatrix}=0$$
Add second and third column on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c &c \\ 
m+n  & x+p &p \\ 
0&  0& 1
\end{vmatrix}=0$$
Laplace on the third line
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c \\ 
m+n  & x+p \\ 
\end{vmatrix}=0$$
Now $a+m+n=p+b+c=k$ then
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\ 
k-a  & x+p \\ 
\end{vmatrix}=0$$
Add first line on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\ 
x+k  & x+k \\ 
\end{vmatrix}=(x-a)(x-p)(x+k)\begin{vmatrix}
x+a & k-p \\ 
1  & 1 \\ 
\end{vmatrix}=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
A: $$ \begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}=0 $$
Subtract the first row to the second row, subtract the third row from the fourth row:
$$ \begin{vmatrix}
x & a & b &c \\ 
a-x & x-a & 0 &0 \\ 
m &n  & x &p \\ 
 0&  0&  p-x& x-p
\end{vmatrix}=0 $$
Factorize $(a-x)$ and $(x-p)$ out:
$$ (a-x)(x-p)\begin{vmatrix}
x & a & b &c \\ 
1 & -1 & 0 &0 \\ 
m &n  & x &p \\ 
 0&  0&  -1& 1
\end{vmatrix}=0 $$
Add the first column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
x & a+x & b &c \\ 
1 & 0 & 0 &0 \\ 
m &n+m  & x &p \\ 
 0&  0&  -1& 1
\end{vmatrix}=0 $$
Compute the determinant by using the second row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b &c \\  
 n+m  & x &p \\ 
   0&  -1& 1
\end{vmatrix}=0 $$
Add the third column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c &c \\  
 n+m  & p+x &p \\ 
   0&  0& 1
\end{vmatrix}=0 $$
Expand the determinant by the last row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c  \\  
 n+m  & p+x  \\ 
\end{vmatrix}=0 $$
Adding the first row to second row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c  \\  
 x+a+n+m  & b+c+p+x  \\ 
\end{vmatrix}=0 $$
Factorize $(x+a+n+m)$ out since $a+m+n=b+c+p$:
$$ (a-x)(x-p)(x+a+n+m)\begin{vmatrix}
  a+x & b+c  \\  
1 & 1  \\ 
\end{vmatrix}=0 $$
$$(a-x)(x-p)(x+a+n+m)(x+a-b-c)=0$$
A: The steps are:
Subtract the first line from the second:
1.Multiply the first by −a−a and add on the second


*Multiply the first by −m−m and add on the third

*Multiply the first by −m−m and add on the 4th
Forth line minus third
Place theorem on the first column
Add second and 3rd column on the second
Place on the third line
Now a+m+n=p+b+c=ka+m+n=p+b+c=k then
Add 1st line on the second
