問題3.3.4
$\left|\begin{array}{ll}a & b \\ b & a\end{array}\right|\left|\begin{array}{ll}c & d \\ d & c\end{array}\right|$ を２通り計算することにより、次の等式を示せ。$$\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=(a c+b d)^{2}-(a d+b c)^{2}$$
ポイント
$$\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\left[\begin{array}{ll}
c & d \\
d & c
\end{array}\right]=\left[\begin{array}{ll}
a c+b d & a d+b c \\
a d+b c & a c+b d
\end{array}\right]$$から示します。
解答例
$$\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\left[\begin{array}{ll}
c & d \\
d & c
\end{array}\right]=\left[\begin{array}{ll}
a c+b d & a d+b c \\
a d+b c & a c+b d
\end{array}\right]$$より、行列式は$$\left|\begin{array}{ll}
a & b \\
b & a
\end{array}\right|\left|\begin{array}{ll}
c & d \\
d & c
\end{array}\right|=\left|\begin{array}{ll}
a c+b d & a d+b c \\
a d+b c & a c+b d
\end{array}\right|$$となる。ここで$$\left|\begin{array}{ll}
a & b \\
b & a
\end{array}\right|\left|\begin{array}{ll}
c & d \\
d & c
\end{array}\right|=\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)$$ $$\left|\begin{array}{ll}
a c+b d & a d+b c \\
a d+b c & a c+b d
\end{array}\right|=(a c+b d)^{2}-(a d+b c)^{2}$$となるから、$$\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=(a c+b d)^{2}-(a d+b c)^{2}$$が成り立つ。
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