# 3元1次方程式の解

$\left\{\begin{array}{l}{a}_{11}{x}_{1}+{a}_{12}{x}_{2}+{a}_{13}{x}_{3}={b}_{1}\text{ }･･････\text{(1)}\\ {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+{a}_{23}{x}_{3}={b}_{2}\text{ }･･････\text{(2)}\\ {a}_{31}{x}_{1}+{a}_{32}{x}_{2}+{a}_{33}{x}_{3}={b}_{3}\text{ }･･････\text{(3)}\end{array}\right\$

${x}_{3}$ を消去する．

(1)×${a}_{23}$
${a}_{11}{a}_{23}{x}_{1}+{a}_{12}{a}_{23}{x}_{2}+{a}_{13}{a}_{23}{x}_{3}={a}_{23}{b}_{1}$ ･･････(4)

(2)×${a}_{13}$
${a}_{13}{a}_{21}{x}_{1}+{a}_{13}{a}_{22}{x}_{2}+{a}_{13}{a}_{23}{x}_{3}={a}_{13}{b}_{2}$ ･･････(5)

(4)−(5)
$\left({a}_{11}{a}_{23}-{a}_{13}{a}_{21}\right){x}_{1}+\left({a}_{12}{a}_{23}-{a}_{13}{a}_{22}\right){x}_{2}={a}_{23}{b}_{1}-{a}_{13}{b}_{2}$ ･･････(6)

(1)×${a}_{33}$
${a}_{11}{a}_{33}{x}_{1}+{a}_{12}{a}_{33}{x}_{2}+{a}_{13}{a}_{33}{x}_{3}={a}_{33}{b}_{1}$･･････(7)

(3)×${a}_{13}$
${a}_{13}{a}_{31}{x}_{1}+{a}_{13}{a}_{32}{x}_{2}+{a}_{13}{a}_{33}{x}_{3}={a}_{13}{b}_{3}$･･････(8)

(7)−(8)
$\left({a}_{11}{a}_{33}-{a}_{13}{a}_{31}\right){x}_{1}+\left({a}_{12}{a}_{33}-{a}_{13}{a}_{32}\right){x}_{2}={a}_{33}{b}_{1}-{a}_{13}{b}_{3}$･･････(9)

${x}_{2}$ を消去する．

(6)×$\left({a}_{12}{a}_{33}-{a}_{13}{a}_{32}\right)$ −(9)× $\left({a}_{12}{a}_{23}-{a}_{13}{a}_{22}\right)$

$\left\{\left({a}_{12}{a}_{33}-{a}_{13}{a}_{32}\right)\left({a}_{11}{a}_{23}-{a}_{13}{a}_{21}\right)-\left({a}_{12}{a}_{23}-{a}_{13}{a}_{22}\right)\left({a}_{11}{a}_{33}-{a}_{13}{a}_{31}\right)\right\}{x}_{1}$$=\left({a}_{12}{a}_{33}-{a}_{13}{a}_{32}\right)\left({a}_{23}{b}_{1}-{a}_{13}{b}_{2}\right)-\left({a}_{12}{a}_{23}-{a}_{13}{a}_{22}\right)\left({a}_{33}{b}_{1}-{a}_{13}{b}_{3}\right)$

${x}_{1}$の係数

${a}_{12}{a}_{33}{a}_{11}{a}_{23}-{a}_{12}{a}_{33}{a}_{13}{a}_{21}$$-{a}_{13}{a}_{32}{a}_{11}{a}_{23}$$+{a}_{13}{a}_{32}{a}_{13}{a}_{21}$$-{a}_{12}{a}_{23}{a}_{11}{a}_{33}$ $+{a}_{12}{a}_{23}{a}_{13}{a}_{31}$$+{a}_{13}{a}_{22}{a}_{11}{a}_{33}-{a}_{13}{a}_{22}{a}_{13}{a}_{31}$

$=-{a}_{12}{a}_{13}{a}_{21}{a}_{33}-{a}_{11}{a}_{13}{a}_{23}{a}_{32}$$+{a}_{13}{a}_{13}{a}_{21}{a}_{32}$$+{a}_{12}{a}_{13}{a}_{23}{a}_{31}$$+{a}_{11}{a}_{13}{a}_{22}{a}_{33}$$-{a}_{13}{a}_{13}{a}_{22}{a}_{31}$

$={a}_{13}\left({a}_{11}{a}_{22}{a}_{33}-{a}_{11}{a}_{23}{a}_{32}$$+{a}_{12}{a}_{23}{a}_{31}$$-{a}_{12}{a}_{21}{a}_{33}$$+{a}_{13}{a}_{21}{a}_{32}-{a}_{13}{a}_{22}{a}_{31}\right)$

${a}_{12}{a}_{33}{a}_{23}{b}_{1}-{a}_{12}{a}_{33}{a}_{13}{b}_{2}$$-{a}_{13}{a}_{32}{a}_{23}{b}_{1}$$+{a}_{13}{a}_{32}{a}_{13}{b}_{2}$$-{a}_{12}{a}_{23}{a}_{33}{b}_{1}$$+{a}_{12}{a}_{23}{a}_{13}{b}_{3}$$+{a}_{13}{a}_{22}{a}_{33}{b}_{1}-{a}_{13}{a}_{22}{a}_{13}{b}_{3}$

$=-{a}_{12}{a}_{13}{a}_{33}{b}_{2}-{a}_{13}{a}_{23}{a}_{32}{b}_{1}$$+{a}_{13}{a}_{13}{a}_{32}{b}_{2}$$+{a}_{12}{a}_{13}{a}_{23}{b}_{3}$$+{a}_{13}{a}_{22}{a}_{33}{b}_{1}$$-{a}_{13}{a}_{13}{a}_{22}{b}_{3}$

$={a}_{13}\left({a}_{22}{a}_{33}{b}_{1}-{a}_{23}{a}_{32}{b}_{1}$$+{a}_{13}{a}_{32}{b}_{2}$$-{a}_{12}{a}_{33}{b}_{2}$$+{a}_{12}{a}_{23}{b}_{3}-{a}_{13}{a}_{22}{b}_{3}\right)$

よって，整理すると

${x}_{1}=\frac{{a}_{13}\left({a}_{22}{a}_{33}{b}_{1}-{a}_{23}{a}_{32}{b}_{1}+{a}_{13}{a}_{32}{b}_{2}-{a}_{12}{a}_{33}{b}_{2}+{a}_{12}{a}_{23}{b}_{3}-{a}_{13}{a}_{22}{b}_{3}\right)}{{a}_{13}\left({a}_{11}{a}_{22}{a}_{33}-{a}_{11}{a}_{23}{a}_{32}+{a}_{12}{a}_{23}{a}_{31}-{a}_{12}{a}_{21}{a}_{33}+{a}_{13}{a}_{21}{a}_{32}-{a}_{13}{a}_{22}{a}_{31}\right)}$

${}_{}=\frac{{b}_{1}{a}_{22}{a}_{33}-{b}_{1}{a}_{23}{a}_{32}+{b}_{2}{a}_{13}{a}_{32}-{b}_{2}{a}_{12}{a}_{33}+{b}_{3}{a}_{12}{a}_{23}-{b}_{3}{a}_{13}{a}_{22}}{{a}_{11}{a}_{22}{a}_{33}-{a}_{11}{a}_{23}{a}_{32}+{a}_{12}{a}_{23}{a}_{31}-{a}_{12}{a}_{21}{a}_{33}+{a}_{13}{a}_{21}{a}_{32}-{a}_{13}{a}_{22}{a}_{31}}$

${x}_{1}=\frac{\left|\begin{array}{ccc}{b}_{1}& {a}_{12}& {a}_{13}\\ {b}_{2}& {a}_{22}& {a}_{23}\\ {b}_{3}& {a}_{32}& {a}_{33}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|}$

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