True/false: $\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$

If you factor $A^2 + I = (A+iI)(A-iI)$ you get $\det(A^2 + I) = \det(A+iI)\cdot\det(A-iI)$. Now I didn't check it till the end, but it seems like $\det(A\pm iI)$ are complex conjugates, so you get something non-negative. $$ \begin{vmatrix} a \pm i & b & c \\ d & e \pm i & f \\ g & h & j \pm i \end{vmatrix} = (a \pm i)(e \pm i)(j \pm i) + \dots = ((ae - 1) \pm i(a+e))(j\pm i) + \dots = (j(ae - 1) - a - e \pm i(ae - 1 + ja + je)) + \dots $$ After the dots it seems trivial.