Let $A$ be a $2 \times 2$ matrix with $\operatorname{det}(A)=-1$ and…

Let $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$, so $adjA=\left|\begin{array}{cc}a& -b\\ -c& d\end{array}\right|$

Now given $\left|A\right|=-1$

So, $ad-bc=-1$

Now $\left|A+I\right|\left|adjA+I\right|=4$

$\Rightarrow \left|AadjA+adjA+A+I\right|=4$

$\Rightarrow \left|\left|A\right|I+adjA+A+I\right|=4$

$\Rightarrow \left|-I+adjA+A+I\right|=4$

$\Rightarrow \left|adjA+A\right|=4$

$\Rightarrow \left|\begin{array}{cc}a+d& 0\\ 0& a+d\end{array}\right|=4$

$\Rightarrow {\left(a+d\right)}^2=4$ 

So, $a+d=2$ or $-2$