Solution: We need to find the trace of $A^2$. If $\lambda _1$ and $\lambda _2$ are the eigenvalues of $A$, then we recall the following:

1. $\lambda _1 + \lambda_2 = \mathop{\mathrm{trace}}(A) $.
2. $\lambda _1\cdot \lambda _2 = \det A$.
3. The eigenvalues of $A^2$ will be $\lambda _1^2$ and $\lambda _2^2$.

Using these facts we have \begin{align*} \lambda _1 + \lambda _2 & = 3, \\ \lambda _1 \cdot \lambda _2 & = 1. \end{align*} We need to find out \[ \mathop{\mathrm{trace}}(A^2) = \lambda _1^2 + \lambda _2^2. \] Note that \begin{align*} & \left( \lambda _1 + \lambda _2 \right)^2 = \lambda _1^2 + \lambda _2^2 + 2\lambda _1\lambda _2 \\ \implies & \lambda _1^2 + \lambda _2^2 = 3^2 - 2 \times 1 = 7. \end{align*} Therefore, \[ \mathop{\mathrm{trace}}\left( A^2 \right) = 7. \]