行列の計算例題（行列式）

例題：次の行列式を計算せよ。

\[ \left|\,\begin{array}{@{}rwr{20pt}wr{20pt}wr{20pt}wr{20pt}@{}}{-}2 & 1 & {-}1 & 0 & {-}2 \\ {-}2 & {-}2 & 0 & 2 & 1 \\ {-}2 & {-}2 & 2 & 2 & 2 \\ 2 & {-}1 & {-}1 & {-}1 & {-}1 \\ 0 & {-}2 & {-}2 & {-}1 & 0\end{array}\,\right| \]

解答

\( \qquad \qquad \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}wr{15pt}@{}}{-}2 & 1 & {-}1 & 0 & {-}2 \\ {-}2 & {-}2 & 0 & 2 & 1 \\ {-}2 & {-}2 & 2 & 2 & 2 \\ 2 & {-}1 & {-}1 & {-}1 & {-}1 \\ 0 & {-}2 & {-}2 & {-}1 & 0\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(0, 1)\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}wr{15pt}@{}}1 & {-}2 & {-}1 & 0 & {-}2 \\ {-}2 & {-}2 & 0 & 2 & 1 \\ {-}2 & {-}2 & 2 & 2 & 2 \\ {-}1 & 2 & {-}1 & {-}1 & {-}1 \\ {-}2 & 0 & {-}2 & {-}1 & 0\end{array}\,\right| \qquad \begin{array}{l}\text{第1列と第2列を}\\\quad\text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}wr{15pt}@{}}1 & {-}2 & {-}1 & 0 & {-}2 \\ 0 & {-}6 & {-}2 & 2 & {-}3 \\ 0 & {-}6 & 0 & 2 & {-}2 \\ 0 & 0 & {-}2 & {-}1 & {-}3 \\ 0 & {-}4 & {-}4 & {-}1 & {-}4\end{array}\,\right|\qquad \begin{array}{l}\text{\((1,1)\)成分を使って}\\\text{第１列の成分を小さくする}\\\qquad[0, -2, -2, -1, -2]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}6 & {-}2 & 2 & {-}3 \\ {-}6 & 0 & 2 & {-}2 \\ 0 & {-}2 & {-}1 & {-}3 \\ {-}4 & {-}4 & {-}1 & {-}4\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}6 & {-}2 & 2 & {-}3 \\ {-}6 & 0 & 2 & {-}2 \\ 0 & {-}2 & {-}1 & {-}3 \\ {-}4 & {-}4 & {-}1 & {-}4\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(2, 2)\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}0 & {-}2 & {-}1 & {-}3 \\ {-}6 & 0 & 2 & {-}2 \\ {-}6 & {-}2 & 2 & {-}3 \\ {-}4 & {-}4 & {-}1 & {-}4\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第3行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}1 & {-}2 & 0 & {-}3 \\ 2 & 0 & {-}6 & {-}2 \\ 2 & {-}2 & {-}6 & {-}3 \\ {-}1 & {-}4 & {-}4 & {-}4\end{array}\,\right| \qquad \begin{array}{l}\text{第1列と第3列を}\\\quad\text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}1 & {-}2 & 0 & {-}3 \\ 0 & {-}4 & {-}6 & {-}8 \\ 0 & {-}6 & {-}6 & {-}9 \\ 0 & {-}2 & {-}4 & {-}1\end{array}\,\right|\qquad \begin{array}{l}\text{\((1,1)\)成分を使って}\\\text{第１列の成分を小さくする}\\\qquad[0, -2, -2, 1]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}4 & {-}6 & {-}8 \\ {-}6 & {-}6 & {-}9 \\ {-}2 & {-}4 & {-}1\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}4 & {-}6 & {-}8 \\ {-}6 & {-}6 & {-}9 \\ {-}2 & {-}4 & {-}1\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(2, 2)\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}2 & {-}4 & {-}1 \\ {-}6 & {-}6 & {-}9 \\ {-}4 & {-}6 & {-}8\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第3行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & {-}4 & {-}2 \\ {-}9 & {-}6 & {-}6 \\ {-}8 & {-}6 & {-}4\end{array}\,\right| \qquad \begin{array}{l}\text{第1列と第3列を}\\\quad\text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & {-}4 & {-}2 \\ 0 & 30 & 12 \\ 0 & 26 & 12\end{array}\,\right|\qquad \begin{array}{l}\text{\((1,1)\)成分を使って}\\\text{第１列の成分を小さくする}\\\qquad[0, 9, 8]\end{array} \)

\( \qquad \qquad \qquad = 1 \times (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}30 & 12 \\ 26 & 12\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}30 & 12 \\ 26 & 12\end{array}\,\right| = (-1) \times ( 30\times 12 - 12\times 26 ) = -48 \)