Let $A$ be a $3\times 3$ matrix such that $A^2=I$ and $A\ne \pm I$. If $\text{tr}(A)=1$, find $\det (A)$.

Solution

1. Since $A^2=I$, the minimal polynomial of $A$ divides $x^2-1=0$, so the eigenvalues $\lambda$ must be $1$ or $-1$.
2. Let the eigenvalues be ${\lambda }_1,{\lambda }_2,{\lambda }_3$. Given $$\begin{array}{rl}\text{tr}(A)& ={\lambda }_1+{\lambda }_2+{\lambda }_3\\ & =1\end{array}$$ and $A\ne \pm I$, the eigenvalues must be $1, 1, -1$.
3. The determinant is the product of the eigenvalues: $$\begin{array}{rl}\det (A)& =(1)(1)(-1)\\ & =-1\end{array}$$.
4. Hence the answer is (B).