Answer
$\det (A^{-1}B^2)^3=\frac{729}{125}$
Work Step by Step
Applying Theorem 3.2.5, we have:
$\det (A^{-1}B^2)^3=\det (A^{-1}B^2.A^{-1}B^2.A^{-1}B^2)=\det (A^{-1}B^2).\det (A^{-1}B^2).\det (A^{-1}B^2)=\det (A^{-1}).\det (B^2).\det (A^{-1}).\det (B^2).\det (A^{-1}).\det (B^2)$
Determinant of $B^T$ can be evaluate by the property P5:
$\det (B^2)=\det (BB)=\det (B).\det (B)=3.3=9$
Applying the property P10, since $\det (A)\ne0$ we get:
$\det (A^{-1})=\frac{1}{\det (A)}=\frac{1}{5}$
Hence, $\det (A^{-1}B^2)^3=\frac{1}{5}.9\frac{1}{5}.9.\frac{1}{5}.9=\frac{729}{125}$
