行列の計算例題（行列式）

例題：次の行列式を計算せよ。

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

解答

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

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

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

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

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

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

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

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}5 & 0 \\ 1 & {-}4\end{array}\,\right| = (-1) \times ( 5\times (-4) - 0\times 1 ) = 20 \)