行列の計算例題（行列式）

　　　　　　　　　　


例題：次の行列式を計算せよ。

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

解答

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

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}1 & 2 & 0 & 2 \\ 0 & {-}5 & 2 & {-}6 \\ 0 & 4 & {-}2 & 0 \\ 0 & 1 & 1 & 2\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, -1, -1]\end{array} \)

\( \qquad \qquad \qquad = 1 \times 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}5 & 2 & {-}6 \\ 4 & {-}2 & 0 \\ 1 & 1 & 2\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}5 & 2 & {-}6 \\ 4 & {-}2 & 0 \\ 1 & 1 & 2\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(2, 0)\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & 1 & 2 \\ 4 & {-}2 & 0 \\ {-}5 & 2 & {-}6\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第3行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & 1 & 2 \\ 0 & {-}6 & {-}8 \\ 0 & 7 & 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, 4, -5]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times 1 \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}6 & {-}8 \\ 7 & 4\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}6 & {-}8 \\ 7 & 4\end{array}\,\right| = (-1) \times ( (-6)\times 4 - (-8)\times 7 ) = -32 \)