Answer
$$A^T=\left[\begin{array}{rrr}{1} & {0} \\ {2} & {1} \\ {-3}& {2}\end{array}\right].$$
$$A^TA= \left[\begin{array}{rrr}{1} & {2} & {-3} \\ {2} & {5} & {-4}\\{-3}&{-4}&{13}\end{array}\right].$$
$$AA^T= \left[\begin{array}{rrr}{14} & {-4} \\ {-4} & {5} \end{array}\right].$$
Work Step by Step
Let $A$ be given by
$$A=\left[\begin{array}{rrr}{1} & {2} & {-3} \\ {0} & {1} & {2}\end{array}\right].$$
Now, we have
$$A^T=\left[\begin{array}{rrr}{1} & {0} \\ {2} & {1} \\ {-3}& {2}\end{array}\right].$$
$$A^TA=\left[\begin{array}{rrr}{1} & {0} \\ {2} & {1} \\ {-3}& {2}\end{array}\right]\left[\begin{array}{rrr}{1} & {2} & {-3} \\ {0} & {1} & {2}\end{array}\right]=\left[\begin{array}{rrr}{1} & {2} & {-3} \\ {2} & {5} & {-4}\\{-3}&{-4}&{13}\end{array}\right].$$
$$AA^T=\left[\begin{array}{rrr}{1} & {2} & {-3} \\ {0} & {1} & {2}\end{array}\right]\left[\begin{array}{rrr}{1} & {0} \\ {2} & {1} \\ {-3}& {2}\end{array}\right]=\left[\begin{array}{rrr}{14} & {-4} \\ {-4} & {5} \end{array}\right].$$