行列の計算例題（行列式）

　　　　　　　　　　


例題：次の行列式を計算せよ。

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

解答

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

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

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

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

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

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

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}1 & 4 & {-}1 \\ 0 & 2 & {-}2 \\ 0 & {-}12 & 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, 0, 2]\end{array} \)

\( \qquad \qquad \qquad = 1 \times 1 \times \left|\,\begin{array}{@{}rwr{15pt}@{}}2 & {-}2 \\ {-}12 & 4\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}@{}}2 & {-}2 \\ {-}12 & 4\end{array}\,\right| = 1 \times ( 2\times 4 - (-2)\times (-12) ) = -16 \)