2元1次方程式の解

${\displaystyle a_{11}{x}_1+a_{12}{x}_2=b_1}$ ･･････(1)

${\displaystyle a_{21}{x}_1+a_{22}{x}_2=b_2}$･･････(2)

${\displaystyle x_2}$を消去する．

(1)×${\displaystyle a_{22}}$
${\displaystyle a_{11}{a}_{22}{x}_1+a_{12}{a}_{22}{x}_2=a_{22}{b}_1}$ ･･････(3)

(2)×${\displaystyle a_{12}}$
${\displaystyle a_{12}{a}_{21}{x}_1+a_{12}{a}_{22}{x}_2=a_{12}{b}_2}$ ･･････(4)

(3)−(4) ${\displaystyle \left(a_{11}{a}_{22}-a_{12}{a}_{21}\right)x_1=a_{22}{b}_1-a_{12}{b}_2}$

${\displaystyle x_1=\frac{a_{22}{b}_1-a_{12}{b}_2}{a_{11}{a}_{22}-a_{12}{a}_{21}}}$

行列式を用いて表すと

${\displaystyle x_1=\frac{\left|\begin{array}{cc}{a}_{22}& b_2\\ a_{12}& b_1\end{array}\right|}{\left|\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right|}=\frac{\left|\begin{array}{cc}{b}_1& a_{12}\\ b_2& a_{22}\end{array}\right|}{\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|}}$

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最終更新日： 2022年10月18日