If $A$ is $2\times2 $ matrix such that $\operatorname{tr} A =\det A=3$ then trace of $A^{-1}=$

Use the following nice and easy formula: $$\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}.$$

The characteristic Polynomial for $A^{-1}$ for any invertible $n \times n$ matrix is $$P_{A^{-1}}(X)=\det(xI-A^{-1})=\det(A^{-1}) \det(xA-I)=x^n\det(A^{-1})\det(A-\frac{1}{x}I)\\=(-1)^nx^n \det(A^{-1}) P_{A}(\frac{1}{x})$$

Now use the fact that for a $2\times 2$ matrix the characteristic polynomial is $$P_B(x)=x^2-\operatorname{tr}(B)x+\det(B)$$

The eigenvalues $\lambda_1$ and $\lambda_2$ are the roots of the characteristic polynomial $\det(A-\lambda 1)=a\lambda^2+b\lambda+c$, where $a,b,c$ are real numbers. It is easy to see that $a=1$. Being the polynomial second degree, the sum of its roots is $\lambda_1+\lambda_2=-b/a=3$ and their product $\lambda_1\cdot\lambda_2=c/a=3$. Then the characteristic polynomial is $\lambda^2-3\lambda+3$. The roots can be computed by Bhaskara and are equal to $\frac{3\pm i\sqrt{3}}{2}$. The trace of $A^{-1}$ is the sum of inverse eigenvalues for $A$: $$ \frac{2}{3+i\sqrt{3}}+\frac{2}{3-i\sqrt{3}}=1. $$