Question
Find the $2 \times 2$ matrix $X$ which satisfies the equation. $\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$
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Answer
${\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]} $
$\Rightarrow\left[\begin{array}{l}0+356+24 \\ 0+204+12\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$\Rightarrow\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$ \Rightarrow 2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] -\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]$
$\Rightarrow 2 X=\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{cc}\frac{-34}{2} \frac{-32}{2} \\ \frac{-24}{2} \frac{-10}{2}\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{ll}-17 & -16 \\ -12 & -5\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}0+356+24 \\ 0+204+12\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$\Rightarrow\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$ \Rightarrow 2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] -\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]$
$\Rightarrow 2 X=\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{cc}\frac{-34}{2} \frac{-32}{2} \\ \frac{-24}{2} \frac{-10}{2}\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{ll}-17 & -16 \\ -12 & -5\end{array}\right]$
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