Step-by-Step Solution
Key Concept: The adjugate matrix scales by $K^{n-1}$ when the original matrix is scaled by $K$, and repeated adjoints follow a predictable power pattern.
Using adjugate matrix properties: $(\text{Adj}(A))^{-1} = \frac{A}{|A|}$, so $(\text{Adj}(KA)) = K^{n-1}\text{Adj}(A)$ for an $n \times n$ matrix. For repeated adjoints, $\text{Adj}(\text{Adj}(KA)) = K^{(n-1)^2}\text{Adj}(\text{Adj}(A)) = K^{(n-1)^2}|A|^{n-2}A$. Also, $A^{-1}\text{Adj}(A^{-1}) = \left[I - \frac{1}{|A|}I\right] = \text{adj}(A^{-1}) = \frac{A}{|A|}$.
Correct Answer: [A-p, q] [B-r] [C-s] [D-p, q]