Answer
$\det (A^2B^5)=6075$
Work Step by Step
Applying Theorem 3.2.5, we have:
$\det (A^2B^5)=\det (A^2).\det (B^5)$
Determinant of $B^T$ can be evaluate by the property P5:
$\det (A^2)=\det (AA)=\det (A).\det (A)=5.5=25$
$\det (B^5)=\det (BBBBB)=\det (B).\det (B).\det (B).\det (B).\det (B)=3.3.3.3.3=243$
Hence, $\det (A^2B^5)=25.243=6075$
