,

จาก $A=\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right]$ จะได้ว่า

$$\begin{eqnarray*} A^{-1}&=&\frac{1}{\det A}\left[\begin{array}{cc}-1&-1\\-1&1\end{array}\right]\\ &=&\frac{1}{(1)(-1)-(1)(1)}\left[\begin{array}{cc}-1&-1\\-1&1\end{array}\right]\\ &=&-\frac12\left[\begin{array}{cc}-1&-1\\-1&1\end{array}\right]\\ &=&\left[\begin{array}{cc}\frac12&\frac12\\\frac12&-\frac12\end{array}\right] \end{eqnarray*}$$

,

จาก $B=\left[\begin{array}{cc} x & y\\ y & z \end{array}\right]$

และ 

$A^{-1}BA=\left[\begin{array}{cc} -2 & 0\\ 0 & 4 \end{array}\right]$

$$\begin{eqnarray*} A^{-1}BA&=&\left[\begin{array}{cc} -2 & 0\\ 0 & 4 \end{array}\right]\\ AA^{-1}BA&=&A\left[\begin{array}{cc} -2 & 0\\ 0 & 4 \end{array}\right]\\ BAA^{-1}&=&A\left[\begin{array}{cc} -2 & 0\\ 0 & 4 \end{array}\right]A^{-1}\\ B&=&\left[\begin{array}{cc}1&1\\1&-1\end{array}\right]\cdot\left[\begin{array}{cc}-2&0\\0&4\end{array}\right]\cdot\left[\begin{array}{cc}\frac12&\frac12\\\frac12&-\frac12\end{array}\right]\\ B&=&\left[\begin{array}{cc}-2&4\\-2&-4\end{array}\right]\cdot\left[\begin{array}{cc}\frac12&\frac12\\\frac12&-\frac12\end{array}\right]\\ B&=&\left[\begin{array}{cc}1&-3\\-3&1\end{array}\right] \end{eqnarray*}$$

,

$$\begin{eqnarray*} B&=&\left[\begin{array}{cc}1&-3\\-3&1\end{array}\right]\\ \left[\begin{array}{cc}x&y\\y&z\end{array}\right]&=&\left[\begin{array}{cc}1&-3\\-3&1\end{array}\right] \end{eqnarray*}$$

ดังนั้น

$$\begin{eqnarray*} x&=&1\\ y&=&-3\\ z&=&1 \end{eqnarray*}$$

จะได้ว่า

$$xyz=-3$$