Given two matrices
[tex]A=\begin{bmatrix}{-18} & {3} & {} \\ {-15} & {-6} & {} \\ {} & {} & {}\end{bmatrix},B=\begin{bmatrix}{-4} & {12} & {} \\ {8} & {-12} & {} \\ {} & {} & {}\end{bmatrix}[/tex]We will solve for the resultant matrix -B - 1/2A.
This operation is represented as
[tex]-B-\frac{1}{2}A=-\begin{bmatrix}{-4} & {12} & {} \\ {8} & {-12} & {} \\ {} & {} & {}\end{bmatrix}-\frac{1}{2}\begin{bmatrix}{-18} & {3} & {} \\ {-15} & {-6} & {} \\ {} & {} & {}\end{bmatrix}[/tex]Let's simplify the matrices further based on scalar operations that can be done here. The B matrix will be multiplied by -1 while the A matrix will be multiplied by 1/2. We now have
[tex]-B-\frac{1}{2}A=\begin{bmatrix}{4} & {-12} & {} \\ {-8} & {12} & {} \\ {} & {} & {}\end{bmatrix}-\begin{bmatrix}{-9} & {\frac{3}{2}} & {} \\ {\frac{-15}{2}} & {-3} & {} \\ {} & {} & {}\end{bmatrix}[/tex]Now, we apply the subtraction of matrices to the simplified matrix operation above. We have
[tex]\begin{gathered} -B-\frac{1}{2}A=\begin{bmatrix}{4-(-9)} & {-12-\frac{3}{2}} & {} \\ {-8-(-\frac{15}{2})} & {12-(-3)} & {} \\ {} & {} & {}\end{bmatrix} \\ -B-\frac{1}{2}A=\begin{bmatrix}{4+9} & {-12-\frac{3}{2}} & {} \\ {-8+\frac{15}{2}} & {12+3} & {} \\ {} & {} & {}\end{bmatrix} \\ -B-\frac{1}{2}A=\begin{bmatrix}{13} & {\frac{-27}{2}} & {} \\ {-\frac{1}{2}} & {15} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex]Hence, the resulting matrix for the operation -B - 1/2A is
[tex]-B-\frac{1}{2}A=\begin{bmatrix}{13} & {\frac{-27}{2}} & {} \\ {-\frac{1}{2}} & {15} & {} \\ {} & {} & {}\end{bmatrix}[/tex]