行列の計算例題（行列式）

　　　　　　　　　　


例題：次の行列式を計算せよ。

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

解答

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

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

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

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

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

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

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

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & {-}6 & 1 \\ 0 & {-}17 & 3 \\ 0 & {-}16 & 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]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}17 & 3 \\ {-}16 & 4\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}17 & 3 \\ {-}16 & 4\end{array}\,\right| = 1 \times ( (-17)\times 4 - 3\times (-16) ) = -20 \)