問題3.5.2a
次の等式を示せ。
（１）$\left|\begin{array}{llll}
1 & 1 & 1 & 1 \\
x & a & a & a \\
x & y & b & b \\
x & y & z & c
\end{array}\right|$
（２）$\left|\begin{array}{llll}
a & b & b & b \\ a & b & a & a \\ a & a & b & a \\ b & b & b & a
\end{array}\right|$
ポイント
ヴァンデルモンドの行列式を利用します。$$\begin{aligned}
& \quad \,\, \left|\begin{array}{cccc}1 & 1 & \cdots & 1 \\ x_{1} & x_{2} & \cdots & x_{n} \\ x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2} \\ \vdots & \vdots & & \vdots \\ x_{1}^{n-1} & x_{2}^{n-1} & \cdots & x_{n}^{n-1}\end{array}\right| \\
&=\prod_{1 \leqq i<j \leqq n}\left(x_{j}-x_{i}\right) \\
&=(-1)^{n(n-1) / 2} \prod_{1 \leqq i<j \leqq n}\left(x_{i}-x_{j}\right) \end{aligned}$$このような形の行列に変形できれば行列式の計算が簡単化できます。
解答例
（１）
$$\begin{array}{l}
& \quad \, \left|\begin{array}{llll}
1 & 1 & 1 & 1 \\
x & a & a & a \\
x & y & b & b \\
x & y & z & c
\end{array}\right| \\
&=\left|\begin{array}{ccccc}
1 & 1 & 1 & 1 \\
0 & a-x & a-x & a-x \\
0 & y-x & b-x & b-x \\
0 & y-x & z-x & c-x
\end{array}\right| \\
&= (a-x)\left|\begin{array}{ccc}
1 & 1 & 1 \\
y-x & b-x & b-x \\
y-x & z-x & c-x
\end{array}\right| \\
&=(a-x)\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & b-y & b-y \\ 0 & z-y & c-y\end{array}\right| \\
&=(a-x)(b-y)\left|\begin{array}{cc}1 & 1 \\ z-y & c-y\end{array}\right| \\
&=(a-x)(b-y)\left|\begin{array}{cc}1 & 1 \\ 0 & c-z\end{array}\right| \\
&=(a-x)(b-y)(c-z) \\
&=-(x-a)(y-b)(z-c) \quad \cdots (\text{答})\end{array}$$
（２）
$$\begin{array}{l}
& \quad \, \left|\begin{array}{llll}
a & b & b & b \\ a & b & a & a \\ a & a & b & a \\ b & b & b & a
\end{array}\right| \\
&=\left|\begin{array}{ccccc}
a & b & b & b \\ 0 & 0 & a-b & a-b \\ 0 & a-b & 0 & a-b \\ b & b & b & a
\end{array}\right| \\
&=(a-b)^{2}\left|\begin{array}{llll}
a & b & b & b \\
0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
b & b & b & a
\end{array}\right| \\
&=(a-b)^{2}\left(a\left|\begin{array}{llll}
0 & 1 & 1 \\
1 & 0 & 1 \\
b & b & a
\end{array}\right|-b\left|\begin{array}{lll}
b & b & b \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|\right)
\end{array}$$
ここで$$\begin{array}{l}& \quad \, \left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ b & b & a\end{array}\right| \\ &=\left|\begin{array}{cc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & a-2 b\end{array}\right| \\ &= \left|\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & a-2 b\end{array}\right| \\ &=-(a-2 b)\end{array}$$および$$\begin{array}{l}& \quad \, \left|\begin{array}{ccc}b & b & b \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right| \\ &=\left|\begin{array}{cc}0 & 0 & -b \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right| \\ &= -\left|\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -b\end{array}\right| \\ &=b\end{array}$$より、$$\begin{array}{l}
& \quad \, (a-b)^{2}\left(a\left|\begin{array}{llll}
0 & 1 & 1 \\
1 & 0 & 1 \\
b & b & a
\end{array}\right|-b\left|\begin{array}{lll}
b & b & b \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array}\right|\right) \\
&=(a-b)^{2}(-a(a-2b)-b^2) \\
&=-(a-b)^4 \quad \cdots (\text{答})
\end{array}$$