問題3.3.1a
次の行列式の値を求めよ。
（１）$\left|\begin{array}{ccc}
5 & -3 & 14 \\
-5 & 6 & 7 \\
10 & 3 & -7
\end{array}\right|$
（２）$\left|\begin{array}{ccc}
2 & 16 & 3 \\
4 & 8 & -6 \\
8 & 8 & 12
\end{array}\right|$
（３）$\left|\begin{array}{cccc}
5 & 4 & 7 & 9 \\
-1 & 3 & 0 & -2 \\
1 & -3 & -8 & 1 \\
5 & 4 & 2 & 11
\end{array}\right|$
（４）$\left|\begin{array}{cccc}
1 & -1 & 2 & 1 \\
2 & -1 & 1 & 2 \\
-1 & 1 & 2 & 1 \\
2 & 1 & 1 & 1
\end{array}\right|$
解法の要点
行列式の値はサラスの方法で求めても良いのですが、計算が煩雑です。簡約化して上三角行列に整理することで行列式を計算しやすくする工夫をしましょう。ブロック行列に分割できる場合はさらに計算を省略できます。
解答例
（１）
$$\begin{aligned}
& \quad \,\, \left|\begin{array}{ccc}
5 & -3 & 14 \\
-5 & 6 & 7 \\
10 & 3 & -7
\end{array}\right| \\
&=5 \cdot 3 \cdot 7\left|\begin{array}{ccc}
1 & -1 & 2 \\
-1 & 2 & 1 \\
2 & 1 & -1
\end{array}\right| \\
&=105\left|\begin{array}{ccc}
1 & -1 & 2 \\
0 & 1 & 3 \\
0 & 3 & -5
\end{array}\right| \begin{array}{l}
\\
②+① \\
③+① \times (-2)
\end{array} \\
&=105\left|\begin{array}{ccc}
1 & -1 & 2 \\
0 & 1 & 3 \\
0 & 0 & -14
\end{array}\right| \begin{array}{l}
\\
\\
③+② \times (-3)
\end{array} \\
&=105 \cdot(-14) \\
&=1470 \quad \cdots (\text{答})
\end{aligned}$$
（２）
$$\begin{aligned}
& \quad \,\, \left|\begin{array}{ccc}
2 & 16 & 3 \\
4 & 8 & -6 \\
8 & 8 & 12
\end{array}\right| \\
&=2 \cdot 8 \cdot 3\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & 1 & -2 \\
4 & 1 & 4 \\
\end{array}\right| \\
&=48\left|\begin{array}{ccccc}
1 & 2 & 1 \\
0 & -3 & -4 \\
0 & -7 & 0
\end{array}\right| \begin{array}{l}
\\
②+① \times (-2) \\
③+① \times (-4)
\end{array} \\
&=366\left|\begin{array}{ccc}
1 & 2 & 1 \\
0 & -3 & -4 \\
0 & -1 & 0 \\
\end{array}\right| \\
&=366\left|\begin{array}{ccccc}
1 & 2 & 1 \\
0 & 0 & -4 \\
0 & -1 & 0
\end{array}\right| \begin{array}{l}
\\
\\
②+③ \times (-3)
\end{array} \\
&=-336\left|\begin{array}{ccc}
1 & 2 & 1 \\
0 & -1 & 0 \\
0 & 0 & -4
\end{array}\right| \\
&= -336 \cdot(-1) \cdot(-4) \\
&= -1344 \quad \cdots (\text{答})
\end{aligned}$$
（３）
$$\begin{aligned}
& \quad \,\, \left|\begin{array}{cccc}
5 & 4 & 7 & 9 \\
-1 & 3 & 0 & -2 \\
1 & -3 & -8 & 1 \\
5 & 4 & 2 & 11
\end{array}\right| \\
&=\left|\begin{array}{cccc}
5 & 4 & 7 & 9 \\
-1 & 3 & 9 & -2 \\
0 & 0 & 1 & -1 \\
0 & 0 & -5 & 2
\end{array}\right| \begin{array}{l}
\\
\\
③+② \\
④+② \times (-1)
\end{array} \\
&=\left|\begin{array}{cc}
5 & 4 \\
-1 & 3
\end{array}\right|\left|\begin{array}{cc}
1 & -1 \\
-5 & 2
\end{array}\right| \\
&=19 \cdot(-3) \\
&=-57 \quad \cdots (\text{答})
\end{aligned}$$
（４）
$$\begin{aligned}
& \quad \,\, \left|\begin{array}{cccc}
1 & -1 & 2 & 1 \\
2 & -1 & 1 & 2 \\
-1 & 1 & 2 & 1 \\
2 & 1 & 1 & 1
\end{array}\right| \\
&=\left|\begin{array}{cccc}
1 & -1 & 2 & 1 \\
0 & 1 & -3 & 0 \\
0 & 0 & 4 & 2 \\
0 & 3 & -3 & -1
\end{array}\right| \begin{array}{l}
\\
②+① \times (-2) \\
③+① \\
④+① \times (-2)
\end{array} \\
&=2\left|\begin{array}{cccc}
1 & -1 & 2 & 1 \\
0 & 1 & -3 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 6 & -1
\end{array}\right| \begin{array}{l}
\\
\\
\\
④+② \times (-3)
\end{array} \\
&=2\left|\begin{array}{cccc}
1 & -1 & 2 & 1 \\
0 & 1 & -3 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & -4
\end{array}\right| \begin{array}{l}
\\
\\
\\
④+③ \times (-2)
\end{array} \\
&=-16 \quad \cdots (\text{答})
\end{aligned}$$