クラメルの公式の導出

${\displaystyle n}$ 元1次連立方程式

${\displaystyle \{\begin{array}{c}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n=b_1\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n=b_2\\ ⋮\\ a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{1n}{x}_n=b_n\end{array}}$　･･････(1)

の解は

${\displaystyle x_j=\frac{\begin{array}{c}\begin{array}{c}\text{line}j\\ \downarrow \end{array}\\ \left|\begin{array}{ccccc}{a}_{11}& \cdots & b_1& \cdots & a_{1n}\\ a_{21}& \cdots & b_2& \cdots & a_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ a_{n1}& \cdots & b_n& \cdots & a_{nn}\end{array}\right|\end{array}}{\left|\begin{array}{ccccc}{a}_{11}& a_{11}& \cdots & \cdots & a_{1n}\\ a_{21}& a_{11}& \cdots & \cdots & a_{2n}\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& & \ddots & ⋮\\ a_{n1}& a_{11}& \cdots & \cdots & a_{nn}\end{array}\right|}}$　･･････(2)

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

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

となる．係数行列 ${\displaystyle \left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right)=A}$とおく．

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

${\displaystyle \left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)=A^{-1}\left(\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right)}$

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

${\displaystyle =\frac{1}{\left|A\right|}\left(\begin{array}{cccc}{\tilde{a}}_{11}& {\tilde{a}}_{21}& \cdots & {\tilde{a}}_{n1}\\ {\tilde{a}}_{12}& {\tilde{a}}_{22}& \cdots & {\tilde{a}}_{n2}\\ ⋮& ⋮& & ⋮\\ {\tilde{a}}_{1n}& {\tilde{a}}_{2n}& \cdots & {\tilde{a}}_{nn}\end{array}\right)\left(\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right)}$

${\displaystyle =\frac{1}{\left|A\right|}\left(\begin{array}{c}{\tilde{a}}_{11}{b}_1+{\tilde{a}}_{21}{b}_2+\cdots +{\tilde{a}}_{n1}{b}_n\\ {\tilde{a}}_{12}{b}_1+{\tilde{a}}_{22}{b}_2+\cdots +{\tilde{a}}_{n2}{b}_n\\ ⋮\\ {\tilde{a}}_{1n}{b}_1+{\tilde{a}}_{2n}{b}_2+\cdots +{\tilde{a}}_{nn}{b}_n\end{array}\right)}$

これより

${\displaystyle x_j=\frac{1}{\left|A\right|}\left({\tilde{a}}_{1j}{b}_1+{\tilde{a}}_{2j}{b}_2+\cdots +{\tilde{a}}_{nj}{b}_n\right)}$　･･････(4)

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

となる．

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

${\displaystyle \left|\begin{array}{ccccc}{a}_{11}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& \cdots & a_{2j}& \cdots & a_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ a_{n1}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right|}$${\displaystyle =a_{1j}{\tilde{a}}_{1j}+a_{2j}{\tilde{a}}_{2j}+\cdots +a_{nj}{\tilde{a}}_{nj}}$　･･････(5)

となる．${\displaystyle A}$の${\displaystyle j}$ 列を${\displaystyle \left(\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right)}$と入れ替えて${\displaystyle j}$ 列で展開すると

${\displaystyle \left|\begin{array}{ccccc}{a}_{11}& \cdots & b_1& \cdots & a_{1n}\\ a_{21}& \cdots & b_2& \cdots & a_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ a_{n1}& \cdots & b_n& \cdots & a_{nn}\end{array}\right|}$${\displaystyle =b_1{\tilde{a}}_{1j}+b_2{\tilde{a}}_{2j}+\cdots +b_n{\tilde{a}}_{nj}}$　･･････(6)

(4)と(6)より

${\displaystyle x_j=\frac{1}{\left|A\right|}\left|\begin{array}{ccccc}{a}_{11}& \cdots & b_1& \cdots & a_{1n}\\ a_{21}& \cdots & b_2& \cdots & a_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ a_{n1}& \cdots & b_n& \cdots & a_{nn}\end{array}\right|}$

${\displaystyle x_j=\frac{\begin{array}{c}\begin{array}{c}\text{line}j\\ \downarrow \end{array}\\ \left|\begin{array}{ccccc}{a}_{11}& \cdots & b_1& \cdots & a_{1n}\\ a_{21}& \cdots & b_2& \cdots & a_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ a_{n1}& \cdots & b_n& \cdots & a_{nn}\end{array}\right|\end{array}}{\left|\begin{array}{ccccc}{a}_{11}& a_{11}& \cdots & \cdots & a_{1n}\\ a_{21}& a_{11}& \cdots & \cdots & a_{2n}\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& & \ddots & ⋮\\ a_{n1}& a_{11}& \cdots & \cdots & a_{nn}\end{array}\right|}}$

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