Given, $A=\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right],$$B=\left[\begin{array}{cc}-3& 2\\ 4& 0\end{array}\right]$ and $C=\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]$ Find the matrix $X$such that $A+2X=2B+C$.

Step 1: We have,

$A=\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right],$ $B=\left[\begin{array}{cc}-3& 2\\ 4& 0\end{array}\right]$ and $C=\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]$

Now, according to the question, we get

$A+2X=2B+C$

Putting the given matrices, we get

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

We have seen that all the given matrices have same order that is $2\times 2$, now

$\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right]+2X=\left[\begin{array}{cc}-6& 4\\ 8& 0\end{array}\right]+\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]$

We know that two matrices have same order then the corresponding elements of both the matrices can be added or subtracted to each other.

$\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right]+2X=\left[\begin{array}{cc}-6+4& 4+0\\ 8+0& 0+2\end{array}\right]$

$\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right]+2X=\left[\begin{array}{cc}-2& 4\\ 8& 2\end{array}\right]$

Step 2 : By using transpose method, we get

$2X=\left[\begin{array}{cc}-2& 4\\ 8& 2\end{array}\right]-\left[\begin{array}{cc}2& -6\\ 2& 0\end{array}\right]$

$\Rightarrow 2X=\left[\begin{array}{cc}-2-2& 4+6\\ 8-2& 2-0\end{array}\right]$

$\Rightarrow 2X=\left[\begin{array}{cc}-4& 10\\ 6& 2\end{array}\right]$

$\Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}-4& 10\\ 6& 2\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{cc}-2& 5\\ 3& 1\end{array}\right]$

Hence, $X=\left[\begin{array}{cc}-2& 5\\ 3& 1\end{array}\right]$