Let's solve for X in the matrix equation first,
[tex]\begin{gathered} 2X+A=-\frac{1}{2}B \\ 2X=-\frac{1}{2}B-A \\ X=\frac{-\frac{1}{2}B-A}{2} \\ or, \\ X=(\frac{1}{2})\times(-\frac{1}{2}B-A) \end{gathered}[/tex]This means, we have to find (-1/2B) and then subtract A.
We then multiply that matrix by the scalar constant (1/2).
*Remember, multiplying a matrix by a scalar means multiplying all the entries of the matrix by that scalar.
* Also, when we subtract matrices, we are basically subtracting each corresponding entry from each other
Let's show the matrix operations:
[tex]\begin{gathered} X=(\frac{1}{2})\times(-\frac{1}{2}B-A) \\ X=(\frac{1}{2})\times(-\frac{1}{2}\begin{bmatrix}{0} & {12} & {} \\ {10} & {-12} & {} \\ {} & {} & {}\end{bmatrix}-\begin{bmatrix}{9} & {-3} & {} \\ {-12} & {9} & {} \\ {} & {} & {}\end{bmatrix}) \\ X=(\frac{1}{2})\times(\begin{bmatrix}0 & -6 \\ -5 & 6\end{bmatrix}-\begin{bmatrix}9 & -3 \\ -12 & 9\end{bmatrix}) \\ X=(\frac{1}{2})\times(\begin{bmatrix}0-9 & -6--3 \\ -5--12 & 6-9\end{bmatrix}) \\ X=(\frac{1}{2})\times\begin{bmatrix}-9 & -3 \\ 7 & -3\end{bmatrix} \\ X=\begin{bmatrix}-\frac{9}{2} & -\frac{3}{2} \\ \frac{7}{2} & -\frac{3}{2}\end{bmatrix} \end{gathered}[/tex]