問題3.4.2
次の連立１次方程式をクラーメルの公式を用いて解け。
（１）$\left[\begin{array}{rrr}1 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]$
（２）$\left[\begin{array}{rrr}2 & -1 & -1 \\ 3 & 1 & 5 \\ 1 & 1 & 3\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c}3 \\ 5 \\ 2\end{array}\right]$
ポイント
$A$が$n$次正則行列であるとき、連立１次方程式$$A\boldsymbol{x}=\boldsymbol{b}$$の解は次のように与えられます。$$x=\left[\begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array}\right], \, x_{1}=\dfrac{\operatorname{det}[a_{1} \cdots \stackrel{i}{b} \cdots a_{n}]}{\operatorname{det}(A)}$$これをクラーメルの公式と言います。ここで$[a_{1} \cdots \stackrel{i}{b} \cdots a_{n}]$は$A$の第 $i$ 例を列ベクトル$\boldsymbol{b}$で置き換えた行列です。
解答例
（１）
$\left[\begin{array}{rrr}1 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]$
$x_{1}=\dfrac{\left|\begin{array}{ccc}0 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right|}{\left|\begin{array}{ccc}1 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right|}=\dfrac{7}{9}$
$x_{2}=\dfrac{\left|\begin{array}{ccc}1 & 0 & 1 \\ 1 & 1 & -1 \\ 2 & 2 & 3\end{array}\right|}{\left|\begin{array}{ccc}1 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right|}=\dfrac{5}{9}$
$x_{3}=\dfrac{\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & 1 & 1 \\ 2 & -1 & 2\end{array}\right|}{\left|\begin{array}{ccc}1 & -2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 3\end{array}\right|}=\dfrac{3}{9}=\dfrac{1}{3}$
$$\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}\dfrac{7}{9} \\ \dfrac{5}{9} \\ \dfrac{1}{3}\end{array}\right] \quad \cdots (\text{答})$$
（２）
$\left[\begin{array}{rrr}2 & -1 & -1 \\ 3 & 1 & 5 \\ 1 & 1 & 3\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c}3 \\ 5 \\ 2\end{array}\right]$
$x_{1}=\dfrac{\left|\begin{array}{ccc}3 & -1 & -1 \\ 5 & 1 & 5 \\ 2 & 1 & 3\end{array}\right|}{\left|\begin{array}{ccc}2 & -1 & -1 \\ 3 & 1 & 5 \\ 1 & 1 & 3\end{array}\right|}=\dfrac{-4}{-2}=2$
$x_{2}=\dfrac{\left|\begin{array}{ccc}2 & 3 & -1 \\ 3 & 5 & 5 \\ 1 & 2 & 3\end{array}\right|}{\left|\begin{array}{ccc}2 & -1 & -1 \\ 3 & 1 & 5 \\ 1 & 1 & 3\end{array}\right|}=\dfrac{-3}{-2}=\frac{3}{2}$
$x_{3}=\dfrac{\left|\begin{array}{ccc}2 & -1 & 3 \\ 3 & 1 & 5 \\ 1 & 1 & 2\end{array}\right|}{\left|\begin{array}{ccc}2 & -1 & -1 \\ 3 & 1 & 5 \\ 1 & 1 & 3\end{array}\right|}=\dfrac{1}{-2}=-\frac{1}{2}$
$$\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c} 2 \\ \dfrac{3}{2} \\ -\dfrac{1}{2}\end{array}\right] \quad \cdots (\text{答})$$