If $A$ is $2\times2 $ matrix such that $\operatorname{tr} A =\det A=3$ then trace of $A^{-1}=$ If $A$ is $2\times2 $ matrix such that $\operatorname{tr} A=\det A=3$ then trace of $A^{-1}$ is?
$(A) \quad 1   \qquad
 (B) \quad \dfrac{1}{3} \qquad
 (C) \quad \dfrac{1}{6} \qquad
 (D) \quad\dfrac{1}{2}$
I did it in this way:
$$\lambda_1+\lambda_2 = 3 $$
$$\lambda_1\cdot\lambda_2=3$$
$$\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \implies
\frac{\lambda_1+\lambda_2}{\lambda_1\lambda_2} \implies
\frac{3}{3}=1$$
I am practicing these types of question and after $5$ minutes of digging I came up with this answer. I wanna know if there is any alternative approach to solve this problem. 
 A: Use the following nice and easy formula:
$$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}
d & -b\\
-c & a
\end{pmatrix}.$$
A: 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 
$$P_B(x)=x^2-\operatorname{tr}(B)x+\det(B)$$
A: 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 $A$:
$$
\frac{2}{3+i\sqrt{3}}+\frac{2}{3-i\sqrt{3}}=1.
$$
A: Every matrix satisfies its own characteristic equation so  
$A$ satisfies  


*

*$A^2-\operatorname{tr}(A)A+\det(A)I=0 $
and   $A^{-1}$ satisfies   

*$A^{-2}-\operatorname{tr}(A^{-1})A^{-1}+\det(A^{-1})I=0$.
With multiplying both sides of the first equation by $A^{-2}$ and dividing by $\det(A)$ we can easily obtain form of the second one.
Edit
I.e.  
$$A^{-2}A^2-\operatorname{tr}(A)A^{-2}A+\det(A)IA^{-2}=0$$
$$I -\operatorname{tr}(A)A^{-1}+\det(A)A^{-2}=0$$
After changing the order of summands and dividing by $\det(A)$
$$A^{-2}- \frac{1}{\det(A)}\operatorname{tr}(A)A^{-1}+\frac{1}{\det(A)}I=0.$$
Comparing the last equation with the second one $\bullet$ we finally obtain   
$$\operatorname{tr}(A^{-1})=\frac{\operatorname{tr}(A)}{\det(A)}$$
A: \begin{align}
\lambda_1 + \lambda_2 = 3 \\
\lambda_1\lambda_2 = 3
\end{align}
So you get $\lambda_2 = \dfrac 3 {\lambda_1},$ so the first equation above becomes
$$
\lambda_1 + \frac 3 \lambda_1 = 3.
$$
Multiply both sides by $\lambda_1$ and you have an ordinary quadratic equation.
The answer posted by "N.S." also leads to the same quadratic equation.
