If A is an invertible matrix of order 2, then $\text{det ((adj A)}^{-1})$ is equal to

$\frac{1}{\text{det A}}$

The correct answer is Option (4) → $\frac{1}{\text{det A}}$

Given: A is an invertible matrix of order 2

Property used: For an invertible matrix $A$ of order $n$,

$\text{adj}(A) = (\det A) \cdot A^{-1}$

$\Rightarrow (\text{adj} A)^{-1} = \frac{1}{\det A} \cdot A$

Therefore,

$\det((\text{adj} A)^{-1}) = \det\left(\frac{1}{\det A} \cdot A\right)$

$= \left(\frac{1}{\det A}\right)^2 \cdot \det A = \frac{1}{\det A}$

Hence, $\det((\text{adj} A)^{-1}) = \frac{1}{\det A}$