We have:
[tex]3A-3B=3\begin{bmatrix}{6} & {3} & {} \\ {-6} & {-3} & {} \\ {4} & {-2} & {}\end{bmatrix}-3\begin{bmatrix}{8} & {-8} & {4} \\ {7} & {-6} & {-5} \\ {} & {} & \end{bmatrix}[/tex]
Multiplying each matrix by 3:
[tex]3A-3B=\begin{bmatrix}{18} & {9} & {} \\ {-18} & {-9} & {} \\ {12} & {-6} & {}\end{bmatrix}-\begin{bmatrix}{24} & {-24} & {12} \\ {21} & {-18} & {-15} \\ {} & {} & \end{bmatrix}[/tex]
Then, to facilitate subtraction, we convert each matrix into a 3x3 matrix, filling in the missing column and row with zero. Like this:
[tex]3A-3B=\begin{bmatrix}{18} & {9} & {0} \\ {-18} & {-9} & {0} \\ {12} & {-6} & {0}\end{bmatrix}-\begin{bmatrix}{24} & {-24} & {12} \\ {21} & {-18} & {-15} \\ {0} & {0} & {0}\end{bmatrix}[/tex]
So, we do the subtraction of each number that makes up the matrix:
[tex]\begin{gathered} 3A-3B=\begin{bmatrix}{18-24} & {9-(-24)} & {0-12} \\ {-18-21} & {-9-(-18)} & {0-(-15)} \\ {12-0} & {-6-0} & {0-0}\end{bmatrix} \\ 3A-3B=\begin{bmatrix}{-6} & {33} & {-12} \\ {-39} & {9} & {15} \\ {12} & {-6} & {0}\end{bmatrix} \end{gathered}[/tex]
Answer:
[tex]\begin{bmatrix}{-6} & {33} & {-12} \\ {-39} & {9} & {15} \\ {12} & {-6} & {0}\end{bmatrix}[/tex]
