Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1  &1  &1  &x \\ 
 1& -1 &1  &1  &y \\ 
 1& 1 & -1 & 1 &z \\ 
 1& 1 & 1 & -1 & u\\ 
 x& y & z & u &0 
\end{vmatrix}= -4 [x^2+y^2+z^2 +u^2 - 2(xy+zx+zu+yz+yu+\color{red}{x}u)]$$
I have tried many things e.g $R_1 \leftrightarrow R_5+ x R_1$ and of course cyclic ($R_2 \leftrightarrow R_5+ y R_2$ etc) and then adding the last row to the second etc. resulting in:
$$\begin{vmatrix}
0 &y+x  &z+x  &u+x  &x^2 \\ 
 y+x&0  &z+y  &u+y  &y^2 \\ 
 x+z& z+y & 0 &z+u  & z^2\\ 
 u+x& u+y &u+z  & 0 &u^2 \\ 
x &y  &z  &u  &0 
\end{vmatrix}$$
but does not look that promising. 
On the other hand if I add the the first column to the others I do get a lot of zeros, since the first row for example takes the form $-1, 0, 0, 0, x+1$. This looks more promising. But in either case I get stuck. 
May I have some hints or answers? 
Edit: The red letters as suggested by Peter.
 A: I think this is meant to be an exercise in the use of symmetries, so let's begin by making a few observations.


*

*Imagine what happens (but don't do it!) when we expand the determinant. All the terms will have exactly one factor from the bottom row, and exactly one factor from the last column. This implies that the determinant is a homogeneous quadratic polynomial $Q(x,y,z,u)$ in the four unknowns. 

*What happens if we exchange the two top rows and the two left most columns? The sign of the determinant gets toggled twice, so remains unchanged. But we also see that $x$ and $y$ trade places - both in the bottom row as well as in the rightmost column. Therefore $Q(x,y,z,u)=Q(y,x,z,u)$. 

*Repeating the previous step for other pairs of rows and columns exactly the same thing happens. Therefore $Q(x,y,z,u)$ is fully symmetric under all the 24 permutations of the unknowns. The conclusion is that
$$
Q(x,y,z,u)=A(x^2+y^2+z^2+u^2)+B(xy+xz+xu+yz+yu+zu)
$$
for some constants $A$ and $B$. This is easier to understand, if you have been exposed to basic properties of elementary symmetric polynomial, but this also follows relatively easily from the observed pairwise symmetries.

*The remaining task is to determine the coefficients $A$ and $B$. If we look at the various ways we can get $x^2$, we see that such terms must include both the SW and NE corners. Imagine expanding along the bottom row. The $x$ in position $(5,1)$ gets a plus sign by the checkerboard rule. It is multiplied by the obvious $4\times$ minor. Imagine expanding that along the right most column. The $x$ is at position $(1,4)$ so gets a minus sign. Altogether we see that the coefficient of the $x^2$ term is the negative of the central $3\times 3$ minor
$$
-\left|\begin{array}{ccc}-1&1&1\\1&-1&1\\1&1&-1\end{array}\right|=-4.
$$
In other words $A=-4$.

*Determining $B$ is slightly trickier, but again by the symmetry it suffices to determine the coefficient of $xy$. The extras come from the fact that we get terms with $xy$ by using either $x$ from the bottom row and $y$ from the rightmost column or the other way around. I leave identifying which two $3\times3$ minors (together with their signs) need to be evaluated to you. They are both equal to four, because they differ from the minor in step 4 only at a position that gets multiplied by a $2\times2$ minor equal to zero. The conclusion will be that $B=8$.

A: Let $Q$ be a real orthogonal matrix whose first column is $(\frac12,\frac12,\frac12,\frac12)^T$. Let also $v=(x,y,z,u)^T$ and $Q^Tv=(a,b,c,d)$. Then the matrix in question is equal to $\pmatrix{Q&0\\ 0&1}A\pmatrix{Q^T&0\\ 0&1}$, where
$$
A=\left[\begin{array}{cccc|c}
2&&&&a\\ &-2&&&b\\ &&-2&&c\\ &&&-2&d\\
\hline
a&b&c&d&0
\end{array}\right].
$$
Hence the required determinant is equal to $\det(A)$. Let $e_1=(1,0,0,0)^T$ and $E\in M_4(\mathbb R)$ be the all-one matrix. By Laplace expansion along the last column of $A$, we see that
$$
\det(A)=8(a^2-b^2-c^2-d^2).
$$
Now we can rewrite this in terms of $x,y,z$ and $u$:
\begin{align}
\det(A)&=8(a^2-b^2-c^2-d^2)\\
&=16a^2-8(a^2+b^2+c^2+d^2)\\
&=16(v^TQe_1)(e_1Q^Tv)-8(v^TQ)(Q^Tv)\\
&=16v^T\left(\frac14E\right)v-8v^Tv\\
&=4(x+y+z+u)^2 - 8(x^2+y^2+z^2+u^2).
\end{align}
A: I am not sure if this is the most efficient way to do it but doing the following row opperations:
$R_2 \to R_2 + R_1, R_3 \to R_3 + R_1, R_4 \to R_4 + R_1, R_5 \to R_5 + xR_1$
Assuming i didnt make any silly mistakes (as is always the case with there types of problems)
$\begin{vmatrix}
-1&1&1&1&x\\
1&-1&1&1&y\\
1&1&-1&1&z\\
1&1&1&-1&u\\
x&y&z&u&0\\
\end{vmatrix}$ $\to$
$\begin{vmatrix}
-1&1&1&1&x\\
0&0&2&2&x+y\\
0&2&0&2&x+z\\
0&2&2&0&x+u\\
0&x+y&x+z&x+u&x^2\\
\end{vmatrix}$ $\to$
$-\begin{vmatrix}
0&2&2&x+y\\
2&0&2&x+z\\
2&2&0&x+u\\
x+y&x+z&x+u&x^2\\
\end{vmatrix}$
Now removing any one of the $2's$ from any row or column and expanding further into a linear combination of $3\times3$ matrices... Havent finished it but this is the most promising version I've tried
