Answer
(a) $|A B|=20$
(b) $|2 A|=32$
(c) the matrices $A$ and $B$ are non-singular because $|A|=4$ and $|B|=5$ (the determinants of the matrices $A$ and $B$ do not equal zero)
(d) $\left|B^{-1}\right|=1/5$ and $\left|A^{-1}\right|=1/4$
(e) $\left|(A B)^{T}\right|=20$
Work Step by Step
Given $|A|=4$ and $|B|=5,$ then
(a) $|A B|=|A||B|=4 * 5=20$
(b) Since the matrix $A$ is a square matrix of order $3,$ then $|2 A|=\left(2^{\wedge} 3\right) *|A|=8 *|A|=8 *$
$4=32$
(c) the matrices $A$ and $B$ are non-singular because $|A|=4$ and $|B|=5$ (the determinants of the matrices $A$ and $B$ do not equal zero)
(d) since the matrices $A$ and $B$ are non-singular, then they are invertible.
So, $A A^{-1}=I$ and then $\left|A A^{-1}\right|=|A|\left|A^{-1}\right|=1$
Then $\left|A^{-1}\right|=\frac{1}{|A|}=\frac{1}{4}$
Also, we have that, $B B^{-1}=I$ and then $\left|B B^{-1}\right|=|B|\left|B^{-1}\right|=1$
Then $\left|B^{-1}\right|=\frac{1}{|B|}=\frac{1}{5}$
(e) $\left|(A B)^{T}\right|=\left|B^{T} A^{T}\right|=\left|B^{T}\right|\left|A^{T}\right|=|B||A|=5 * 4=20$
