Finding the determinant of a matrix I am given $A = \begin{pmatrix}
a & b\\ 
c & d
\end{pmatrix} $ and B = $ \begin{pmatrix}
e & f\\ 
g & h
\end{pmatrix}$ whose elements are non-zero reals. 

If $BA = I$, where $I$ is the $2 \times 2$ identity matrix and $D$ is the value of the determinant of $B$, then find the value of $D$

Assume that four options are given for the correct answer (which is $\frac{d}{e}$) and only one is correct. How can I find the correct answer quickly?
ADDED:
The answer suggested in my module is $\frac{d}{e}$, so I am suppose to derive to that point.
 A: The determinant of $B$ is given by $eh-gf$ (see here). Presumably, the question is asking you to describe this quantity in terms of $a,b,c,d$. In that case, here are some facts you can combine to find the answer:


*

*$\det(XY)=\det(X)\det(Y)$ for all matrices $X$ and $Y$ 

*$\det(I)=1$

*$\det(A)=ad-bc$
These facts together show that $D=\det(B)=\frac{1}{ad-bc}$. Now note that the entries of $B=A^{-1}$, in terms of the entries of $A$, are (see here)
$$B=\begin{pmatrix}\frac{d}{ad-bc}\,\,\,\, & \frac{-b}{ad-bc}\\ \stackrel{\vphantom{g}}{\frac{-a}{ad-bc}}\,\,\,\, & \frac{c}{ad-bc}\end{pmatrix}$$
Thus, $\displaystyle e=\frac{d}{ad-bc}$, so that $\displaystyle\frac{d}{e}=ad-bc$, which is wrong. The correct answer is $\displaystyle\frac{e}{d}=\frac{1}{ad-bc}$.
A: HINT $\: $ Multiplicative maps $\rm\:d\:$ preserve products so inverses $\rm\ A\:B = 1\ \Rightarrow\ d(A)\:d(B) = d(1) = 1\:.\ $   
Note: $\rm\ 1^2 = 1\ \Rightarrow\ d(1)^2 = d(1)\ $ so $\rm\ d(1) = 1\ $ if the target is a domain and $\rm\ d\not\equiv 0\:.$
