Question
If $\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right]+3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]$ Find $A$

Answer

$\begin{array}{l}{\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right]+3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]} \end{array} $
$ 3 A=\left[\begin{array}{cc}-2 & -2 \\ 1 & -3\end{array}\right]-\left[\begin{array}{cc}4 & -2 \\ 4 & 0\end{array}\right]  $
$ 3 A=\left[\begin{array}{cc}-2-4-2+2 \\ 1-4 & -3-0\end{array}\right]  $
$ 3 A=\left[\begin{array}{cc}-6 & 0 \\ -3-3\end{array}\right]  $
$ A=\frac{1}{3}\left[\begin{array}{cc}-6 & 0 \\ -3 & -3\end{array}\right]=\left[\begin{array}{cc}-2 & 0 \\ -1-1\end{array}\right]$

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