Question:

Step 1: Use the formula $adj(adj(A)) = (\det A)^{n-2} A$ for $n\times n$ matrix. For $n=3$: $adj(adj(A)) = (\det A)^1 \cdot A = (\det A)A$. Step 2: Apply $adj$ again: $adj(adj(adj(A))) = adj((\det A)\cdot A) = (\det A)^2 \cdot adj(A)$. Step 3: So $(\det A)^2 \cdot adj(A) = \begin{bmatrix}16 & 0 & 4\\ 5 & 4 & 0\\ 1 & 4 & 3\end{bmatrix}$. Step 4: Compute determinant of the given matrix $M = \begin{bmatrix}16 & 0 & 4\\ 5 & 4 & 0\\ 1 & 4 & 3\end{bmatrix}$: $$\det(M) = 16(12-0) - 0 + 4(20-4) = 192 + 64 = 256$$ Step 5: Since $\det((\det A)^2 \cdot adj(A)) = (\det A)^6 \cdot (\det A)^2 = (\det A)^8 = 256 = 2^8$, so $\det A = 2$ (positive). Step 6: Then $(\det A)^2 = 4$, so $adj(A) = \frac{M}{4} = \begin{bmatrix}4 & 0 & 1\\ 5/4 & 1 & 0\\ 1/4 & 1 & 3/4\end{bmatrix}$. Step 7: $\det(adj\ A) = (\det A)^{n-1} = 2^2 = 4$. ✓ Option (3) is correct. Option (4) is incorrect. Step 8: $A^{-1} = \frac{adj(A)}{\det A} = \frac{adj(A)}{2}$. Trace$(A^{-1}) = \frac{1}{2}\text{trace}(adj(A)) = \frac{1}{2}\cdot\frac{1}{4}\text{trace}(M) = \frac{1}{8}(16+4+3) = \frac{23}{8}$. So $8\cdot\text{trace}(A^{-1}) = 23$. ✓ Option (1) is correct.