Let $A$ be a \(2 \times~ 2\) matrix with det \((A) = -1\) and det \(((A + I) ~(\)Adj \((A) + I)) = 4\). Then the sum of the diagonal elements of $A$ can be

\(\left |(A + I)(\textrm{adj }A + I) \right | = 4\)

\(\Rightarrow~\left | A ~\textrm{adj }A + A + \textrm{adj } A + I\right | = 4\)

\(\Rightarrow~\left | (A)|I + A + \textrm{adj } A + I \right | = 4\)

\(\left | A\right | = ~–1 \Rightarrow~\left | A + \textrm{adj } A \right | = 4\)

\(A=\begin{bmatrix}
a&b \\
c&d
\end{bmatrix}~\textrm{adj}~A=\begin{bmatrix}
a&-b \\
-c&d
\end{bmatrix}\)

\(\Rightarrow~ \begin{bmatrix}
(a+d)&0 \\
0&(a+d)
\end{bmatrix}=4\)

\(\Rightarrow ~a+d=\pm~2\)