# クラメルの公式

## ■2元1次連立方程式の場合

$\left\{\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}\right\$

を行列を用いて，

$\left(\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right)$

と表す．

$A=\left(\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right)$とおく．

$\left|A\right|=\left|\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right|\ne 0$のとき，連立方程式の解は，

$x=\frac{1}{\left|A\right|}\left|\begin{array}{cc}{c}_{1}& {b}_{1}\\ {c}_{2}& {b}_{2}\end{array}\right|=\frac{\left|\begin{array}{cc}{c}_{1}& {b}_{1}\\ {c}_{2}& {b}_{2}\end{array}\right|}{\left|\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right|}$

$y=\frac{1}{\left|A\right|}\left|\begin{array}{cc}{a}_{1}& {c}_{1}\\ {a}_{2}& {c}_{2}\end{array}\right|=\frac{\left|\begin{array}{cc}{a}_{1}& {c}_{1}\\ {a}_{2}& {c}_{2}\end{array}\right|}{\left|\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right|}$

で与えられる．

これらの解を表す式をクラメルの公式という．

## ■3元1次連立方程式の場合

$\left\{\begin{array}{c}{a}_{1}x+{b}_{2}y+{c}_{1}z={d}_{1}\\ {a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2}\\ {a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3}\end{array}\right\$

を行列を用いて

$\left(\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}{d}_{1}\\ {d}_{2}\\ {d}_{3}\end{array}\right)$

と表す．

$A=\left(\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right)$とおく．

$\left|A\right|=\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right|\ne 0$ のとき，連立方程式の解は，

$x=\frac{1}{\left|A\right|}\left|\begin{array}{ccc}{d}_{1}& {b}_{1}& {c}_{1}\\ {d}_{2}& {b}_{2}& {c}_{2}\\ {d}_{3}& {b}_{3}& {c}_{3}\end{array}\right|=\frac{\left|\begin{array}{ccc}{d}_{1}& {b}_{1}& {c}_{1}\\ {d}_{2}& {b}_{2}& {c}_{2}\\ {d}_{3}& {b}_{3}& {c}_{3}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right|}$

$y=\frac{1}{\left|A\right|}\left|\begin{array}{ccc}{a}_{1}& {d}_{1}& {c}_{1}\\ {a}_{2}& {d}_{2}& {c}_{2}\\ {a}_{3}& {d}_{3}& {c}_{3}\end{array}\right|=\frac{\left|\begin{array}{ccc}{a}_{1}& {d}_{1}& {c}_{1}\\ {a}_{2}& {d}_{2}& {c}_{2}\\ {a}_{3}& {d}_{3}& {c}_{3}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right|}$

$z=\frac{1}{\left|A\right|}\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {d}_{1}\\ {a}_{2}& {b}_{2}& {d}_{2}\\ {a}_{3}& {b}_{3}& {d}_{3}\end{array}\right|=\frac{\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {d}_{1}\\ {a}_{2}& {b}_{2}& {d}_{2}\\ {a}_{3}& {b}_{3}& {d}_{3}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}\end{array}\right|}$

で与えられる．

## ■n元1次連立方程式の場合

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

を行列を用いて

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

と表す．

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

$|A|=|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{12}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|\ne 0$のとき，連立方程式の解は，

${x}_{1}=\frac{1}{|A|}|\begin{array}{cccc}{b}_{1}& {a}_{12}& \cdots & {a}_{1n}\\ {b}_{2}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|=\frac{|\begin{array}{cccc}{b}_{1}& {a}_{12}& \cdots & {a}_{1n}\\ {b}_{2}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {b}_{n}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|}{|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|}$

${x}_{2}=\frac{1}{|A|}|\begin{array}{cccc}{a}_{11}& {b}_{1}& \cdots & {a}_{1n}\\ {a}_{21}& {b}_{2}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {b}_{n}& \cdots & {a}_{nn}\end{array}|=\frac{|\begin{array}{cccc}{a}_{11}& {b}_{1}& \cdots & {a}_{1n}\\ {a}_{21}& {b}_{2}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {b}_{n}& \cdots & {a}_{nn}\end{array}|}{|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|}$

$⋮$

${x}_{n}=\frac{1}{|A|}|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {b}_{1}\\ {a}_{21}& {a}_{22}& \cdots & {b}_{2}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {b}_{n}\end{array}|=\frac{|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {b}_{1}\\ {a}_{21}& {a}_{22}& \cdots & {b}_{2}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {b}_{n}\end{array}|}{|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|}$

で与えられる．導出

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