Let A be a 3×3 real matrix with A^2=A and tr(A)=2. Then det (I+A) equals

If $\lambda$ is an eigenvalue of $A$ then $\lambda^2=\lambda$, so every eigenvalue is $0$ or $1$. The trace is the sum of the three eigenvalues (with multiplicity), so exactly two of them equal $1$ and one equals $0$. The eigenvalues of $I+A$ are $1+\lambda$, namely $2,2,1$, and the determinant is their product, $4$. As a check, $A=\operatorname{diag}(1,1,0)$ gives $\det(I+A)=2\cdot2\cdot1=4$. The value $\det(2I-A)=2$ is a different quantity.