# クラメルの公式の導出

$n$ 元1次連立方程式

$\left\{\begin{array}{c}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}+\cdots +{a}_{1n}{x}_{n}={b}_{1}\\ {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+\cdots +{a}_{2n}{x}_{n}={b}_{2}\\ ⋮\\ {a}_{n1}{x}_{1}+{a}_{n2}{x}_{2}+\cdots +{a}_{1n}{x}_{n}={b}_{n}\end{array}$　･･････(1)

の解は

${x}_{j}=\frac{\begin{array}{c}\begin{array}{c}\text{line}\text{ }j\\ ↓\end{array}\\ \left|\begin{array}{ccccc}{a}_{11}& \cdots & {b}_{1}& \cdots & {a}_{1n}\\ {a}_{21}& \cdots & {b}_{2}& \cdots & {a}_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ {a}_{n1}& \cdots & {b}_{n}& \cdots & {a}_{nn}\end{array}\right|\end{array}}{\left|\begin{array}{ccccc}{a}_{11}& {a}_{11}& \cdots & \cdots & {a}_{1n}\\ {a}_{21}& {a}_{11}& \cdots & \cdots & {a}_{2n}\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& & \ddots & ⋮\\ {a}_{n1}& {a}_{11}& \cdots & \cdots & {a}_{nn}\end{array}\right|}$　･･････(2)

( $j=1,2,\cdots ,n$ )

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

$\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{n}\end{array}\right)=\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ ⋮\\ {b}_{n}\end{array}\right)$　･･････(3)

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

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

$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{n}\end{array}\right)={A}^{-1}\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ ⋮\\ {b}_{n}\end{array}\right)$

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

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

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

これより

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

( $j=1,2,\cdots ,n$ )

となる．

$A$ $j$ 列で展開すると

$\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|$$={a}_{1j}{\stackrel{˜}{a}}_{1j}+{a}_{2j}{\stackrel{˜}{a}}_{2j}+\cdots +{a}_{nj}{\stackrel{˜}{a}}_{nj}$　･･････(5)

となる．$A$$j$ 列を$\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ ⋮\\ {b}_{n}\end{array}\right)$と入れ替えて$j$ 列で展開すると

$\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|$$={b}_{1}{\stackrel{˜}{a}}_{1j}+{b}_{2}{\stackrel{˜}{a}}_{2j}+\cdots +{b}_{n}{\stackrel{˜}{a}}_{nj}$　･･････(6)

(4)と(6)より

${x}_{j}=\frac{1}{\left|A\right|}\left|\begin{array}{ccccc}{a}_{11}& \cdots & {b}_{1}& \cdots & {a}_{1n}\\ {a}_{21}& \cdots & {b}_{2}& \cdots & {a}_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ {a}_{n1}& \cdots & {b}_{n}& \cdots & {a}_{nn}\end{array}\right|$

${x}_{j}=\frac{\begin{array}{c}\begin{array}{c}\text{line}\text{ }j\\ ↓\end{array}\\ \left|\begin{array}{ccccc}{a}_{11}& \cdots & {b}_{1}& \cdots & {a}_{1n}\\ {a}_{21}& \cdots & {b}_{2}& \cdots & {a}_{2n}\\ ⋮& & ⋮& & ⋮\\ ⋮& & ⋮& & ⋮\\ {a}_{n1}& \cdots & {b}_{n}& \cdots & {a}_{nn}\end{array}\right|\end{array}}{\left|\begin{array}{ccccc}{a}_{11}& {a}_{11}& \cdots & \cdots & {a}_{1n}\\ {a}_{21}& {a}_{11}& \cdots & \cdots & {a}_{2n}\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& & \ddots & ⋮\\ {a}_{n1}& {a}_{11}& \cdots & \cdots & {a}_{nn}\end{array}\right|}$

となり，公式が得られる．

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