Answer
$$AD-DA=\left[\begin{array}{r}{3}&{-6} \\ {3}&{-3} \end{array}\right].$$
Work Step by Step
Since $$
A=\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right], \quad D=\left[\begin{array}{r}{0}&{-3} \\ {-2}&{1} \end{array}\right]
$$
we have
$$
AD=\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]\left[\begin{array}{r}{0}&{-3} \\ {-2}&{1} \end{array}\right]=
\left[\begin{array}{r}{3}&{-15} \\ {-7}&{5} \end{array}\right]$$
$$
DA=\left[\begin{array}{r}{0}&{-3} \\ {-2}&{1} \end{array}\right]\left[\begin{array}{ll}{3} & {0}\\
{-1}&{5}\end{array}\right]=
\left[\begin{array}{r}{0} & {-9}\\
{-10}&{8} \end{array}\right]$$
It is easy to note that
$$AD-DA=\left[\begin{array}{r}{3}&{-15} \\ {-7}&{5} \end{array}\right]-\left[\begin{array}{r}{0} & {-9}\\
{-10}&{8} \end{array}\right]=\left[\begin{array}{r}{3}&{-6} \\ {3}&{-3} \end{array}\right].$$
