問題2.1.1
次の連立1次方程式を掃き出し法で解け。
（１）$\left\{\begin{align}
2 x_{1}+3 x_{2}&=-1 \\
x_{1}-x_{2}&=2
\end{align}\right.$
（２）$\left\{\begin{align}
3 x_{1}+2 x_{2}&=0 \\
x_{1}-2 x_{2}&=8
\end{align}\right.$
（３）$\left\{\begin{align}
x_{1}+2 x_{2}-x_{3}&=2 \\
-x_{1}+3 x_{3}&=8 \\
x_{2}-2 x_{3}&=-4
\end{align}\right.$
（４）$\left\{\begin{align}
x_{1}+x_{2}-x_{3}&=1 \\
2 x_{1}+x_{2}+3 x_{3}&=4 \\
-x_{1}+2 x_{2}-4 x_{3}&=-2
\end{align}\right.$
ポイント
基本変形を行って連立1次方程式を解く方法を掃き出し法と言います。掃き出し法のやり方については教科書を参照して下さい。
以下では$x$をあらわに書いていますが、行列の形で計算した方が書く手間が大きく省けます。皆さんが実際に問題を解く際は行列の形で基本変形していきましょう。
解答例
（１）
$\left\{\begin{aligned} 2 x_{1}+3 x_{2} &=-1 \\ x_{1}-x_{2} &=2 \end{aligned}\right.$
$\left\{\begin{aligned} 5 x_{2} &=-5 \quad ①+② \times(-2)\\ x_{1}-x_{2} &=2 \end{aligned}\right.$
$\left\{\begin{aligned} x_{2} &=-1 \quad ① \times \dfrac{1}{5} \\ x_{1}-x_{2} &=2 \end{aligned}\right.$
$\left\{\begin{aligned} x_{1}-x_{2} &=2 \\ x_{2} &=-1 \end{aligned}\right.$
$\left\{\begin{aligned} x_{1} &=1 \\ x_{2} &=-1 \end{aligned}\right. \quad \cdots (\text{答})$
（２）
$\left\{\begin{aligned} 3 x_{1}+2 x_{2} &=0 \\ x_{1}-2 x_{2} &=8 \end{aligned}\right.$
$\left\{\begin{aligned} 8 x_{2} &=-24 \quad ①+②\times(-3)\\ x_{1}-2 x_{2} &=8 \end{aligned}\right.$
$\left\{\begin{aligned} x_{2} &=-3 \quad ① \times \dfrac{1}{8} \\ x_{1}-2 x_{2} &=8 \end{aligned}\right.$
$\left\{\begin{aligned} & & x_{2} &=-3 \\ x_{1} & & &=2 \quad \quad ②+① \times 2 \end{aligned}\right.$
$\left\{\begin{aligned} x_{1} &=2 \\ x_{2} &=-3 \end{aligned}\right. \quad \cdots (\text{答})$
（３）
$\left\{\begin{array}{rrrl}x_{1}&+2 x_{2}&-x_{3}&=2 \\ -x_{1}& &+3 x_{3}&=8 \\ & x_{2} &-2 x_{3}&=-4\end{array}\right.$
$\left\{\begin{aligned} x_{1}+2 x_{2}-x_{3} &=2 \\ 2 x_{2}+2 x_{3} &=10 \quad ②+① \\ x_{2}-2 x_{3} &=-4 \end{aligned}\right.$
$\left\{\begin{array}{rrrl}x_{1}&+2 x_{2}&-x_{3}&=2 \\ &x_{2}&+x_{3}&=5 \quad ② \times \dfrac{1}{2}\\ &x_{2}&-2 x_{3}&=-4\end{array}\right.$
$\left\{\begin{array}{rrrl} x_{1} & &-3 x_{3}& =-8 \quad ①+② \times(-2) \\ &x_{2}&+x_{3} &=5 \\ & &-3 x_{3} &=-9 \quad ③-② \end{array}\right.$
$\left\{\begin{array}{rrrl} x_{1}& &-3 x_{3}&=-8 \\ &x_{2}&+x_{3} &=5 \\ & & x_{3} &=3 \quad ③ \times \left(-\dfrac{1}{3}\right) \end{array}\right.$
$\left\{\begin{array}{rrrl} x_{1}& & &=1 \quad ①+③ \times 3 \\ & x_{2}& &=2 \quad ②+③ \times(-1)\\ & & x_{3} &=3 \end{array}\right.$
$\left\{\begin{aligned} x_{1} &=1 \\ x_{2} &=2 \\ x_{3} &=3 \end{aligned}\right. \quad \cdots (\text{答})$
（４）
$\left\{\begin{array}{rrrl}x_{1}&+x_{2}&-x_{3}&=1 \\ 2 x_{1}&+x_{2}&+3 x_{3}&=4 \\ -x_{1}&+2 x_{2}&-4 x_{3}&=-2\end{array}\right.$
$\left\{\begin{array}{rrrll}x_{1}&+x_{2}&-x_{3}&= 1 \\ &-x_{2}&+5 x_{3}&=2 & ②+① \times(-2)\\ & 3 x_{2}&-5 x_{3}&=-1 & ③+① \end{array}\right.$
$\left\{\begin{array}{rrrll}x_{1} &&+\quad 4 x_{3}&=3 & ①+② \\ &-x_{2}&+5 x_{3}&=2 \\ &&10 x_{3}&=5 & ③+② \times 3\end{array}\right.$
$\left\{\begin{array}{rrrll}x_{1} & & +4 x_{3} &=3 \\ &x_{2}&-5 x_{3}&=-2 & ② \times(-1) \\ & &x_{3} &=\dfrac{1}{2} & ③ \times \dfrac{1}{10} \end{array}\right.$
$\left\{\begin{array}{rrrll} x_{1} & & &=1 & ①+③ \times (-4) \\ & x_{2} & &=\dfrac{1}{2} & ②+③ \times 5 \\ & & x_{3} &= \dfrac{1}{2} \end{array}\right.$
$\left\{\begin{aligned} x_{1} &=1 \\ x_{2} &=\dfrac{1}{2} \\ x_{3} &=\dfrac{1}{2} \end{aligned}\right. \quad \cdots (\text{答})$