# 基本的な行列の問題

## ■問題

$\left|\begin{array}{cccc}3& 2& 8& -1\\ 4& -5& -2& 3\\ 1& 3& 2& 4\\ -5& 7& 5& 1\end{array}\right|$

## ■答

$702$

## ■計算

$\left|\begin{array}{cccc}3& 2& 8& -1\\ 4& -5& -2& 3\\ 1& 3& 2& 4\\ -5& 7& 5& 1\end{array}\right|$

$=2×{\left(-1\right)}^{1+2}\left|\begin{array}{ccc}4& -2& 3\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$ $-5×{\left(-1\right)}^{2+2}\left|\begin{array}{ccc}3& 8& -1\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$ $+3×{\left(-1\right)}^{3+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ -5& 5& 1\end{array}\right|$ $+7×{\left(-1\right)}^{4+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ 1& 2& 4\end{array}\right|$　 ･･･（1）

$2×{\left(-1\right)}^{1+2}\left|\begin{array}{ccc}4& -2& 3\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$

この計算においても行列式の値は第2列で展開(余因子展開)して求めることにする．

$=-2\left\{-2×{\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 4\\ -5& 1\end{array}\right|+2×{\left(-1\right)}^{2+2}\left|\begin{array}{cc}4& 3\\ -5& 1\end{array}\right|+5×{\left(-1\right)}^{3+2}\left|\begin{array}{cc}4& 3\\ 1& 4\end{array}\right|\right\}$

$=-2\left\{2\left(1+20\right)+2\left(4+15\right)-5\left(16-3\right)\right\}$

$=-2\left(42+38-65\right)$

$=-30$　 ･･･（2）

$-5×{\left(-1\right)}^{2+2}\left|\begin{array}{ccc}3& 8& -1\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$

$=-5\left\{8{×\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 4\\ -5& 1\end{array}\right|+2×{\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& -1\\ -5& 1\end{array}\right|+5×{\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& -1\\ 1& 4\end{array}\right|\right\}$

$=-5\left\{-8\left(1+20\right)+2\left(3-5\right)-5\left(12+1\right)\right\}$

$=-5\left(-168-4-65\right)$

$=1185$　 ･･･（3）

$3×{\left(-1\right)}^{3+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ -5& 5& 1\end{array}\right|$

$=-3\left\{8{×\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& 3\\ -5& 1\end{array}\right|-2×{\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& -1\\ -5& 1\end{array}\right|+5×{\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& -1\\ 4& 3\end{array}\right|\right\}$

$=-3\left\{-8\left(4+15\right)-2\left(3-5\right)-5\left(9+4\right)\right\}$

$=-3\left(-152+4-65\right)$

$=639$　 ･･･（4）

$7×{\left(-1\right)}^{4+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ 1& 2& 4\end{array}\right|$

$=7\left\{8{×\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& 3\\ 1& 4\end{array}\right|-2×{\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& -1\\ 1& 4\end{array}\right|+2×{\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& -1\\ 4& 3\end{array}\right|\right\}$

$=7\left\{-8\left(16-3\right)-2\left(12+1\right)-2\left(9+4\right)\right\}$

$=7\left(-104-26-26\right)$

$=-1092$　 ･･･（5）

（1）に（2）〜（5）をそれぞれ代入する．

$2×{\left(-1\right)}^{1+2}\left|\begin{array}{ccc}4& -2& 3\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$ $-5×{\left(-1\right)}^{2+2}\left|\begin{array}{ccc}3& 8& -1\\ 1& 2& 4\\ -5& 5& 1\end{array}\right|$ $+3×{\left(-1\right)}^{3+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ -5& 5& 1\end{array}\right|$ $+7×{\left(-1\right)}^{4+2}\left|\begin{array}{ccc}3& 8& -1\\ 4& -2& 3\\ 1& 2& 4\end{array}\right|$

$=-30+1185+639-1092$

$=702$

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