問題1.4.2
次の行列の方程式と同等な連立1次方程式を求めよ。
（１）$$\left[\begin{array}{rrr}
2 & 1 & 3 \\
0 & -1 & 2 \\
1 & 0 & -1
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right]=\left[\begin{array}{r}
1 \\
2 \\
-2
\end{array}\right]$$
（２）$$\left[\begin{array}{rrr}
3 & 0 & 1 \\
1 & -1 & 2
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right]=\left[\begin{array}{r}
-1 \\
0
\end{array}\right]$$
ポイント
$x_{1}$、$x_{2}$、$\cdots$、$x_{n}$ を未知数とする連立1次方程式$$\left\{\begin{array}{c}
a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}=b_{1} \\
a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}=b_{2} \\
\vdots \\
a_{m 1} x_{1}+a_{m 2} x_{2}+\cdots+a_{m n} x_{n}=b_{n}
\end{array}\right.$$は、以下の行列の方程式$$\left[\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n}
\end{array}\right]\left[\begin{array}{c}
x_{1} \\
x_{2} \\
\vdots \\
x_{n}
\end{array}\right]=\left[\begin{array}{c}
b_{1} \\
b_{2} \\
\vdots \\
b_{m}
\end{array}\right]$$と同等です。
解答例
（１）
$$\left\{\begin{array}{l}
2 x_{1}+x_{2}+3 x_{3}=1 \\
-x_{2}+2 x_{3}=2 \\
x_{1}-x_{3}=-2
\end{array}\right. \quad \cdots (\text{答})$$
（２）
$$\left\{\begin{array}{l}
3 x_{1}+ x_{3}=-1 \\
x_{1}-x_{2}+2 x_{3}=0
\end{array}\right. \quad \cdots (\text{答})$$