Answer
$$\left[\begin{array}{ccc}{-\frac{2}{3}} & {\frac{4}{3}} \\ {\frac{10}{3}} & {4}\end{array}\right].$$
Work Step by Step
Let $$
A=\left[\begin{array}{ccc}{1} & {2} \\ {3} & {4}\end{array}\right], \quad B=\left[\begin{array}{ccc}{-1} & {0} \\ {1} & {1}\end{array}\right]
$$
We have
$$3X=2(A+2B)\Longrightarrow X=\frac{2}{3}(A+2B)=\frac{2}{3}A+\frac{4}{3}B$$
and hence $$X=\frac{2}{3}\left[\begin{array}{ccc}{1} & {2} \\ {3} & {4}\end{array}\right]+\frac{4}{3}\left[\begin{array}{ccc}{-1} & {0} \\ {1} & {1}\end{array}\right]=\left[\begin{array}{ccc}{-\frac{2}{3}} & {\frac{4}{3}} \\ {\frac{10}{3}} & {4}\end{array}\right].$$
