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(B) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. Given $\det(A) = ad - bc = -1$.We know that $\operatorname{Adj}(A) = \begin{bmatrix} d & -

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$2$

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(B) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$. Given $\det(A) = ad - bc = -1$.We know that $\operatorname{Adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Then $A + I = \begin{bmatrix} a+1 & b \ c & d+1 \end{bmatrix}$ and $\operatorname{Adj}(A) + I = \begin{bmatrix} d+1 & -b \ -c & a+1 \end{bmatrix}$.Calculating the product $(A+I)(\operatorname{Adj}(A)+I)$:$(A+I)(\operatorname{Adj}(A)+I) = \begin{bmatrix} a+1 & b \ c & d+1 \end{bmatrix} \begin{bmatrix} d+1 & -b \ -c & a+1 \end{bmatrix} = \begin{bmatrix} (a+1)(d+1)-bc & -(a+1)b + b(a+1) \ c(d+1) - c(d+1) & -bc + (d+1)(a+1) \end{bmatrix}$.This simplifies to $\begin{bmatrix} ad+a+d+1-bc & 0 \ 0 & ad+a+d+1-bc \end{bmatrix}$.Since $ad-bc = -1$,the diagonal elements are $a+d+1-1 = a+d$.Thus,the matrix is $\begin{bmatrix} a+d & 0 \ 0 & a+d \end{bmatrix}$.The determinant is $(a+d)^2 = 4$.Therefore,$a+d = 2$ or $a+d = -2$.Comparing with the options,the sum of the diagonal elements (trace of $A$) can be $2$.

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

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