Solution: Since $\det (A + \iota I) = 0$, so $-\iota $ is an eigenvalue of $A$. As $A$ is a real matrix and hence the characteristic polynomial will be of real coefficient. Therefore, complex roots will appear in pair and hence, $\iota $ will also be an eigenvalue. If $\lambda $ is the third eigenvalue of $A$, then \begin{align*} \det A = 3 & \implies \lambda (\iota )(-\iota ) = 3 \implies \lambda = 3. \end{align*} Thus, eigenvalues of $A^2$ will be $\iota ^2 = -1 = (-\iota )^2$ and $3^2 = 9$ and hence, \[ \operatorname{trace}\left( A^2 \right) = \text{ sum of the eigenvalues of } A^2 = 7. \]