TITLE: If $A$ is $2\times2 $ matrix such that $\operatorname{tr} A =\det A=3$ then trace of $A^{-1}=$
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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   
$$\operatorname{tr}(A^{-1})=\frac{\operatorname{tr}(A)}{\det(A)}$$