Let A be a 2 × 2 matrix with det(A)=-1 and det((A+I)(Adj(A)+I))=4 ....

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

Now given \left|A\right| = - 1$</p><p>So,$ad - bc = - 1$</p><p>Now \left|A + I\right|\left|adj A + I\right| = 4$

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

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

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

$⇒ \left|adjA + A\right| = 4$

⇒ \left|\begin{cases}a + d & 0 \\ 0 & a + d\end{cases}\right| = 4

$⇒ {\left(a + d\right)}^{2} = 4$

So, $a + d = 2$ or $- 2$