クラメルの公式の導出

$3$\displaystyle{n}$ 元1次連立方程式

$3$\displaystyle{ \{ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{x}_{1}+\hspace{1}{a}_{ 12 }\hspace{1}{x}_{2}+\cdots +\hspace{1}{a}_{ 1n }\hspace{1}{x}_{n}=\hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{x}_{1}+\hspace{1}{a}_{ 22 }\hspace{1}{x}_{2}+\cdots +\hspace{1}{a}_{ 2n }\hspace{1}{x}_{n}=\hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 }\hspace{1}{x}_{1}+\hspace{1}{a}_{ n2 }\hspace{1}{x}_{2}+\cdots +\hspace{1}{a}_{ 1n }\hspace{1}{x}_{n}=\hspace{1}{b}_{n} & \vspace{6}\\ \end{array} }$ 　･･････(1)

の解は

$3$\displaystyle{\hspace{1}{x}_{j}=\frac{\hspace{2} \begin{array} \begin{array} \text{line}\text{ }j & \vspace{6}\\ ↓& \vspace{6}\\ \end{array} & \vspace{6}\\ \left|{ \begin{array} \hspace{1}{a}_{ 11 } &\cdots & \hspace{1}{b}_{1} &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } &\cdots & \hspace{1}{b}_{2} &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } &\cdots & \hspace{1}{b}_{n} &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} }\right| & \vspace{6}\\ \end{array} \hspace{2}}{\hspace{2} \left|{ \begin{array} \hspace{1}{a}_{ 11 } & \hspace{1}{a}_{ 11 } &\cdots &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } & \hspace{1}{a}_{ 11 } &\cdots &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots &\vdots &\ddots & &\vdots & \vspace{6}\\ \vdots &\vdots & &\ddots &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } & \hspace{1}{a}_{ 11 } &\cdots &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} }\right| \hspace{2}}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
　･･････(2)

( $3$\displaystyle{j=1,2,\cdots ,n}$ )

となる．(2)の解の式のことをクラメルの公式という．

■導出

(1)を行列を用いて表すと

$3$\displaystyle{ $ \begin{array} \hspace{1}{a}_{ 11 } & \hspace{1}{a}_{ 12 } &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } & \hspace{1}{a}_{ 22 } &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } & \hspace{1}{a}_{ n2 } &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} $ $ \begin{array} \hspace{1}{x}_{1} & \vspace{6}\\ \hspace{1}{x}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{x}_{n} & \vspace{6}\\ \end{array} $ = $ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} $ }$ 　･･････(3)

となる．係数行列 $3$\displaystyle{ $ \begin{array} \hspace{1}{a}_{ 11 } & \hspace{1}{a}_{ 12 } &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } & \hspace{1}{a}_{ 22 } &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } & \hspace{1}{a}_{ n2 } &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} $ =A}$ とおく．

$3$\displaystyle{ \|A\| \neq 0}$ のとき， $3$\displaystyle{A}$ は逆行列 $3$\displaystyle{{A}^{ - 1 }}$ が存在し，(3)の両辺に左から $3$\displaystyle{{A}^{ - 1 }}$ をかけると

$3$\displaystyle{\left({ \begin{array} \hspace{1}{x}_{1} & \vspace{6}\\ \hspace{1}{x}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{x}_{n} & \vspace{6}\\ \end{array} }\right)={A}^{ - 1 }\left({ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{{A}^{ - 1 }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left|{A}\right| \hspace{2}}\left({ \begin{array} \hspace{1}{\tilde{a}}_{ 11 } & \hspace{1}{\tilde{a}}_{ 21 } &\cdots & \hspace{1}{\tilde{a}}_{ n1 } & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 12 } & \hspace{1}{\tilde{a}}_{ 22 } &\cdots & \hspace{1}{\tilde{a}}_{ n2 } & \vspace{6}\\ \vdots &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 1n } & \hspace{1}{\tilde{a}}_{ 2n } &\cdots & \hspace{1}{\tilde{a}}_{ nn } & \vspace{6}\\ \end{array} }\right)}$ 　(ここを参照)より

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left|{A}\right| \hspace{2}}\left({ \begin{array} \hspace{1}{\tilde{a}}_{ 11 } & \hspace{1}{\tilde{a}}_{ 21 } &\cdots & \hspace{1}{\tilde{a}}_{ n1 } & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 12 } & \hspace{1}{\tilde{a}}_{ 22 } &\cdots & \hspace{1}{\tilde{a}}_{ n2 } & \vspace{6}\\ \vdots &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 1n } & \hspace{1}{\tilde{a}}_{ 2n } &\cdots & \hspace{1}{\tilde{a}}_{ nn } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left|{A}\right| \hspace{2}}\left({ \begin{array} \hspace{1}{\tilde{a}}_{ 11 }\hspace{1}{b}_{1}+\hspace{1}{\tilde{a}}_{ 21 }\hspace{1}{b}_{2}+\cdots +\hspace{1}{\tilde{a}}_{ n1 }\hspace{1}{b}_{n} & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 12 }\hspace{1}{b}_{1}+\hspace{1}{\tilde{a}}_{ 22 }\hspace{1}{b}_{2}+\cdots +\hspace{1}{\tilde{a}}_{ n2 }\hspace{1}{b}_{n} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{\tilde{a}}_{ 1n }\hspace{1}{b}_{1}+\hspace{1}{\tilde{a}}_{ 2n }\hspace{1}{b}_{2}+\cdots +\hspace{1}{\tilde{a}}_{ nn }\hspace{1}{b}_{n} & \vspace{6}\\ \end{array} }\right)}$

これより

$3$\displaystyle{\hspace{1}{x}_{j}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left|{A}\right| \hspace{2}}\left({ \hspace{1}{\tilde{a}}_{ 1j }\hspace{1}{b}_{1}+\hspace{1}{\tilde{a}}_{ 2j }\hspace{1}{b}_{2}+\cdots +\hspace{1}{\tilde{a}}_{ nj }\hspace{1}{b}_{n} }\right)}$ 　･･････(4)

( $3$\displaystyle{j=1,2,\cdots ,n}$ )

となる．

$3$\displaystyle{A}$ を $3$\displaystyle{j}$ 列で展開すると

$3$\displaystyle{\left|{ \begin{array} \hspace{1}{a}_{ 11 } &\cdots & \hspace{1}{a}_{ 1j } &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } &\cdots & \hspace{1}{a}_{ 2j } &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } &\cdots & \hspace{1}{a}_{ nj } &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} }\right|}$ $3$\displaystyle{=\hspace{1}{a}_{ 1j }\hspace{1}{\tilde{a}}_{ 1j }+\hspace{1}{a}_{ 2j }\hspace{1}{\tilde{a}}_{ 2j }+\cdots +\hspace{1}{a}_{ nj }\hspace{1}{\tilde{a}}_{ nj }}$ 　･･････(5)

となる． $3$\displaystyle{A}$ の $3$\displaystyle{j}$ 列を $3$\displaystyle{\left({ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} }\right)}$ と入れ替えて $3$\displaystyle{j}$ 列で展開すると

$3$\displaystyle{\left|{ \begin{array} \hspace{1}{a}_{ 11 } &\cdots & \hspace{1}{b}_{1} &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } &\cdots & \hspace{1}{b}_{2} &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } &\cdots & \hspace{1}{b}_{n} &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} }\right|}$ $3$\displaystyle{=\hspace{1}{b}_{1}\hspace{1}{\tilde{a}}_{ 1j }+\hspace{1}{b}_{2}\hspace{1}{\tilde{a}}_{ 2j }+\cdots +\hspace{1}{b}_{n}\hspace{1}{\tilde{a}}_{ nj }}$ 　･･････(6)

(4)と(6)より

$3$\displaystyle{\hspace{1}{x}_{j}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left|{A}\right| \hspace{2}}\left|{ \begin{array} \hspace{1}{a}_{ 11 } &\cdots & \hspace{1}{b}_{1} &\cdots & \hspace{1}{a}_{ 1n } & \vspace{6}\\ \hspace{1}{a}_{ 21 } &\cdots & \hspace{1}{b}_{2} &\cdots & \hspace{1}{a}_{ 2n } & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \vdots & &\vdots & &\vdots & \vspace{6}\\ \hspace{1}{a}_{ n1 } &\cdots & \hspace{1}{b}_{n} &\cdots & \hspace{1}{a}_{ nn } & \vspace{6}\\ \end{array} }\right|}$

となり，公式が得られる．

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最終更新日：2022年9月9日