Question
Solve the following equations for $X$ and $Y$, if $3 X-Y==\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $X-3 Y=$
$\left[\begin{array}{ll}0 & -1 \\ 0 & -1\end{array}\right]$
$\left[\begin{array}{ll}0 & -1 \\ 0 & -1\end{array}\right]$
$=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$
(i)
and $X-3 Y=\left[\begin{array}{ll}0 & -1 \\ 0 & -1\end{array}\right] \ldots \ldots \ldots \ldots \ldots$ (ii)
By (i) x 3 – (ii) we get
$8 X=3\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$
$=\left[\begin{array}{cc}3 & -3 \\ -3 & 3\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$
$=\left[\begin{array}{cc}3-0 & -3+1 \\ -3-0 & 3+1\end{array}\right]$
$\therefore \quad 8 X=\left[\begin{array}{cc}3 & -2 \\ -3 & 4\end{array}\right]$
$\therefore \quad X=\frac{1}{8}\left[\begin{array}{cc}3 & -2 \\ -3 & 4\end{array}\right]$
$\therefore \quad X=\left[\begin{array}{cc}\frac{3}{8} & \frac{-2}{8} \\ \frac{-3}{8} & \frac{4}{8}\end{array}\right]=\left[\begin{array}{cc}\frac{3}{8} & \frac{-1}{4} \\ \frac{-3}{8} & \frac{1}{2}\end{array}\right]$
By (i) - (ii) $\times 3$, we get
$8 Y=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-3\left[\begin{array}{cc}0 & -1 \\ 0 & -1\end{array}\right]$
$=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -3 \\ 0 & -3\end{array}\right]$
$=\left[\begin{array}{cc}1-0 & -1+3 \\ -1-0 & 1+3\end{array}\right]$
$\therefore \quad 8 Y=\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]$
$\therefore \quad Y=\frac{1}{8}\left[\begin{array}{cc}1 & 2 \\ -1 & 4\end{array}\right]=\left[\begin{array}{cc}\frac{1}{8} & \frac{2}{8} \\ \frac{-1}{8} & \frac{4}{8}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{8} & \frac{1}{4} \\ \frac{-1}{8} & \frac{1}{2}\end{array}\right]$
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