Answer:
[tex]\large\boxed{\left[\begin{array}{ccc}-18&-33\\-42&-27\end{array}\right] }[/tex]
Step-by-step explanation:
[tex]n\cdot\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =\left[\begin{array}{ccc}(n)(a)&(n)(b)\\(n)(c)&(n)(d)\end{array}\right]\\\\============================\\\\3\cdot\left[\begin{array}{ccc}-6&-11\\-14&-9\end{array}\right] =\left[\begin{array}{ccc}(3)(-6)&(3)(-11)\\(3)(-14)&(3)(-9)\end{array}\right] \\\\=\left[\begin{array}{ccc}-18&-33\\-42&-27\end{array}\right][/tex]
