$3 \times 3$ real matrix: relation with determinants

Question

$A$ is a $3 \times 3$ matrix with real entries such that $\operatorname{det}(A+I_3)=\operatorname{det}(A+2I_3)$. Then is $2\operatorname{det}(A+I_3)+\operatorname{det}(A-I_3)+ 6 =3 \operatorname{det}A$? So, if the first holds, then does the second also hold? That $6$ in the sum makes it somehow "different". Any idea / solution?

Answer 1

Let $\lambda_{1,2,3}$ denote the eigenvalues of $A$, then $\operatorname{det}(A+I)=\operatorname{det}(A+2I)$ can be rewritten as $$(\lambda_1+1)(\lambda_2+1)(\lambda_3+1)=(\lambda_1+2)(\lambda_2+2)(\lambda_3+2),$$ which is in turn equivalent to (check it!) $$ \lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3+3(\lambda_1+\lambda_2+\lambda_3)+7=0.\tag{1}$$ On the other hand, \begin{align} 2\operatorname{det}(A+I)+\operatorname{det}(A-I)-3\operatorname{det}A+6=\\ =2(\lambda_1+1)(\lambda_2+1)(\lambda_3+1)+(\lambda_1-1)(\lambda_2-1)(\lambda_3-1)-3\lambda_1\lambda_2\lambda_3+6=\\ =\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3+3(\lambda_1+\lambda_2+\lambda_3)+7,\;\; \end{align} and this vanishes according to (1).