If A is an invertible matrix of order 2, then $\text{det ((adj A)}^{-1})$ is equal to
Answer & explanation
Correct answer: option 4
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, $\det((\text{adj} A)^{-1}) = \frac{1}{\det A}$