If A is an invertible matrix of order 2, then det (A−1) is equal to

Since $A$ is an invertible matrix, $A^{-1}$ exists and $A^{-1}=\frac{1}{|A|}$ adjA.

As matrix $A$ is of order 2, let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$.

Then, $|A|=a d-b c$ and adj $A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$.

Now,

$A^{-1}=\frac{1}{|A|}$ adjA $=\left[\begin{array}{cc}\frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|}\end{array}\right]$

$\therefore\left|A^{-1}\right|=\left|\begin{array}{cc}\frac{d}{|A|} & \frac{-b}{|A|} \\ \frac{-c}{|A|} & \frac{a}{|A|} \mid\end{array}\right|=\frac{1}{|A|^{2}}\left|\begin{array}{cc}d & -b \\ -c & a\end{array}\right|=\frac{1}{|A|^{2}}(a d-b c)=\frac{1}{|A|^{2}} \cdot|A|=\frac{1}{|A|}$

$\therefore \operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}$

Hence, the correct answer is B.