Answer
$\det ((B)^{-1}(2A)B^{T})=80$
Work Step by Step
applying the property $P9$, we obtain:
$\det ((B)^{-1}(2A)B^{T})=\det (B)^{-1}.\det (2A).\det(B^{T})$
Since $\det (B) \ne 0$ we have:
$\det ((B)^{-1}=\frac{1}{\det (B)}=\frac{1}{3}$
Since $A$ is $4 \times 4$ matrix, hence we get:
$\det (2A)=2^4\det (A)=16.5=80$
Using property $P5$, $\det ((B)^T=\det (B)=3$
Hence here, $\det ((B)^{-1}(2A)B^{T})=\frac{1}{3}.80.3=80$
