行列の計算例題（行列式）

例題：次の行列式を計算せよ。

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

解答

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

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

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}wr{15pt}@{}}1 & {-}2 & 2 & 0 & 0 \\ 0 & 5 & {-}4 & {-}1 & {-}2 \\ 0 & 6 & {-}6 & 0 & 0 \\ 0 & {-}3 & 2 & 2 & {-}1 \\ 0 & 5 & {-}5 & 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, 2, -1, 2]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}5 & {-}4 & {-}1 & {-}2 \\ 6 & {-}6 & 0 & 0 \\ {-}3 & 2 & 2 & {-}1 \\ 5 & {-}5 & 1 & {-}1\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}5 & {-}4 & {-}1 & {-}2 \\ 6 & {-}6 & 0 & 0 \\ {-}3 & 2 & 2 & {-}1 \\ 5 & {-}5 & 1 & {-}1\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(0, 2)\end{array} \)

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

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}wr{15pt}@{}}{-}1 & {-}4 & 5 & {-}2 \\ 0 & {-}6 & 6 & 0 \\ 0 & {-}6 & 7 & {-}5 \\ 0 & {-}9 & 10 & {-}3\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, -1]\end{array} \)

\( \qquad \qquad \qquad = 1 \times (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}6 & 6 & 0 \\ {-}6 & 7 & {-}5 \\ {-}9 & 10 & {-}3\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}6 & 6 & 0 \\ {-}6 & 7 & {-}5 \\ {-}9 & 10 & {-}3\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(2, 2)\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}9 & 10 & {-}3 \\ {-}6 & 7 & {-}5 \\ {-}6 & 6 & 0\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第3行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}3 & 10 & {-}9 \\ {-}5 & 7 & {-}6 \\ 0 & 6 & {-}6\end{array}\,\right| \qquad \begin{array}{l}\text{第1列と第3列を}\\\quad\text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}3 & 10 & {-}9 \\ {-}2 & {-}3 & 3 \\ 0 & 6 & {-}6\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, 1, 0]\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}3 & 10 & {-}9 \\ {-}2 & {-}3 & 3 \\ 0 & 6 & {-}6\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(1, 0)\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}2 & {-}3 & 3 \\ {-}3 & 10 & {-}9 \\ 0 & 6 & {-}6\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第2行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}2 & {-}3 & 3 \\ {-}1 & 13 & {-}12 \\ 0 & 6 & {-}6\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, 1, 0]\end{array} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}2 & {-}3 & 3 \\ {-}1 & 13 & {-}12 \\ 0 & 6 & {-}6\end{array}\,\right|\qquad \begin{array}{l}\text{非零（絶対値）最小元を探す}\\\qquad(1, 0)\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 13 & {-}12 \\ {-}2 & {-}3 & 3 \\ 0 & 6 & {-}6\end{array}\,\right| \qquad \begin{array}{l}\text{第1行と第2行を}\\\quad \text{入れ替える}\end{array} \)

\( \qquad \qquad \qquad = (-1) \times \left|\,\begin{array}{@{}rwr{15pt}wr{15pt}@{}}{-}1 & 13 & {-}12 \\ 0 & {-}29 & 27 \\ 0 & 6 & {-}6\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} \)

\( \qquad \qquad \qquad = (-1) \times (-1) \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}29 & 27 \\ 6 & {-}6\end{array}\,\right|\qquad\text{段を減らす} \)

\( \qquad \qquad \qquad = 1 \times \left|\,\begin{array}{@{}rwr{15pt}@{}}{-}29 & 27 \\ 6 & {-}6\end{array}\,\right| = 1 \times ( (-29)\times (-6) - 27\times 6 ) = 12 \)