行列式の計算

\[\begin{align}
\begin{array}{|ccccc|}
0 & 1 & x & x & 1 \\
1 & 0 & 1 & x & x \\
x & 1 & 0 & 1 & x \\
x & x & 1 & 0 & 1 \\
1 & x & x & 1 & 0 \\
\end{array}
&=
\begin{array}{|ccccc|}
2(x + 1) & 2(x + 1) & 2(x + 1) & 2(x + 1) & 2(x + 1) \\
1 & 0 & 1 & x & x \\
x & 1 & 0 & 1 & x \\
x & x & 1 & 0 & 1 \\
1 & x & x & 1 & 0 \\
\end{array} \\
&=
\begin{array}{|c:cccc|}
2(x + 1) & 0 & 0 & 0 & 0 \\ \hdashline
1 & -1 & 0 & x – 1 & x – 1 \\
x & 1-x & -x & 1-x & 0 \\
x & 0 & 1-x & -x & 1-x \\
1 & x-1 & x-1 & 0 & -1 \\
\end{array} \\
&= 2 (x + 1)
\begin{array}{|c:ccc|}
-1 & 0 & 0 & \\ \hdashline
1-x & -x & x(1-x) & -(1-x)^2 \\
0 & 1-x & -x & 1-x \\
x-1 & x-1 & (x-1)^2 & x(x-2) \\
\end{array} \\
&= – 2(x + 1)
\begin{array}{|ccc|}
-x & x(1-x) & -(1-x)^2 \\
0 & x^2 – 3x + 1 & x^2 -3x + 1 \\
x -1 & (x-1)^2 & x(x-2) \\
\end{array} \\
&= -2(x+1)(x^2 – 3x + 1)
\begin{array}{|ccc|}
-x & x-x^2 & -(x-1)^2 \\
0 & 1 & 1 \\
x-1 & (x-1)^2 & x^2 – 2 x \\
\end{array} \\
&= – 2(x+1)(x^2-3x+1)
\begin{array}{|c:c:c|}
-x & -x^2 + x & x-1 \\ \hdashline
0 & 1 & 0 \\ \hdashline
x-1 & (x-1)^2 & -1 \\
\end{array} \\
&= -2(x+1)(x^2-3x+1)
\begin{array}{|cc|}
-x & x-1 \\
x-1 & -1 \\
\end{array} \\
&=2(x+1)(x^2-3x+1)^2 \\
&=2(x+1)\left(x – \frac{3 + \sqrt{5}}{2}\right)^2 \left(x – \frac{3 – \sqrt{5}}{2}\right)^2
\end{align}\]