第18章 18.7

$pa + qb + rc = x$ を行列で表すと

$\begin{pmatrix}2 & 1 & 1 \\ 0 & 1 & -1 \\ -1 & -1 & 1\end{pmatrix}\begin{pmatrix}p \\ q \\ r\end{pmatrix} = \begin{pmatrix}x \\ y \\ z\end{pmatrix}$

$\left[\begin{array}{ccc|c} 2 & 1 & 1 & x \\ 0 & 1 & -1 & y \\ -1 & -1 & 1 & z \end{array}\right] \to \left[\begin{array}{ccc|c} 2 & 0 & 2 & x-y \\ 0 & 1 & -1 & y \\ -1 & 0 & 0 & y+z \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 0 & 0 & -y-z \\ 0 & 1 & -1 & y \\ 2 & 0 & 2 & x-y \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 0 & 0 & -y-z \\ 0 & 1 & -1 & y \\ 0 & 0 & 2 & x+y+2z \end{array}\right]$

$\to \left[\begin{array}{ccc|c} 1 & 0 & 0 & -y-z \\ 0 & 1 & -1 & y \\ 0 & 0 & 1 & \frac{x+y+2z}{2} \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & 0 & 0 & -y-z \\ 0 & 1 & 0 & \frac{x+3y+2z}{2} \\ 0 & 0 & 1 & \frac{x+y+2z}{2} \end{array}\right]$

これから

$p = -y-z, \quad q = \frac{1}{2}x + \frac{3}{2}y + z, \quad r = \frac{1}{2}x + \frac{1}{2}y + z$