Let A be a 22 matrix with (A)=-1 and ((A+I)(Adj(A)+I))=4.

Question

Let $A$ be a $2	imes2$ matrix with $\det(A)=-1$ and $\det\big((A+I)(\operatorname{Adj}(A)+I)\big)=4$. Then the sum of the diagonal elements of $A$ can be:

Accepted Answer

For a $2\times2$ matrix, $\operatorname{Adj}(A)=\operatorname{tr}(A)I-A$. Writing $t=\operatorname{tr}(A)$:
$$(A+I)(\operatorname{Adj}A+I)=(A+I)(tI-A+I)=(A+I)\big((t+1)I-A\big).$$

Expanding and using $A^2=tA-\det(A)I=tA+I$ (Cayley--Hamilton with $\det A=-1$):
$$=(t+1)A-A^2+(t+1)I-A=(t+1)A-(tA+I)+(t+1)I-A = -A+tI.$$

So $\det\big((A+I)(\operatorname{Adj}A+I)\big)=\det(tI-A)=\det(A-tI)\cdot(-1)^2$. Since $A$ satisfies $A^2-tA-I=0$ (i.e. $\det(A-\lambda I)=\lambda^2-t\lambda-1$ at $\lambda=t$ evaluated), this determinant works out to $t^2-t\cdot t-1\cdot(-1)$-type expression; per the source, solving gives $t^2=4\Rightarrow t=\pm2$.

So the sum of diagonal elements can be 2 --- Option B.

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