Answer
$$(I_2-D)^2=\left[\begin{array}{r}{7}&{3} \\ {2}&{6} \end{array}\right].$$
Work Step by Step
Let $$
D=\left[\begin{array}{r}{0}&{-3} \\ {-2}&{1} \end{array}\right]
$$
we have
$$I_2-D=D=\left[\begin{array}{r}{1}&{0} \\ {0}&{1} \end{array}\right]-\left[\begin{array}{r}{0}&{-3} \\ {-2}&{1} \end{array}\right]=\left[\begin{array}{r}{0}&{3} \\ {2}&{0} \end{array}\right].$$
Hence, $$(I_2-D)^2=\left[\begin{array}{r}{0}&{3} \\ {2}&{0} \end{array}\right].\left[\begin{array}{r}{0}&{3} \\ {2}&{0} \end{array}\right]=\left[\begin{array}{r}{7}&{3} \\ {2}&{6} \end{array}\right].$$
