Q:

Suppose $A$ is a 2 x 2 matrix, prove that if $A^{3} = 0$, then $A^{2} = 0$

Suppose $A$ is a 2 x 2 matrix, prove that if $A^{3} = 0$, then $A^{2} = 0$
I was given this equation as a hint:
$$A^{2} - tr(A)A + det(A)I = 0$$
where $tr(A)$ is the sum of the diagonal entries of $A$. 
My attempt:
$$A^{2} - tr(A)A = -det(A)I$$
$$A(A - tr(A)) = -det(A)I$$
$$A[-\frac{1}{det(A)}A(A - tr(A))] = I$$
$A$ is invertible, thus $det(A)$ is not equal to $0$. Multiplying $A^{2}$ to both sides of equation 2:
$$A^{3}(A - tr(A)) = A^{2}(-det(A)I)$$ 
Since $A^{3} = 0$ and since $-det(A)I$ is not equal to $0$, then $A^{2}$ must be equal.  QED. 
Is this proof correct? 
EDIT: There seems to be a similar question, but my question is more specific in that I have to use the equation given in the hint. 

A:

If $A^3=0$, $\det{A^3} = (\det{A})^3 = 0$, so the equation becomes
$$ A^2 -A\operatorname{tr}{A} = 0.  \tag{1} $$
Multiplying by $A$,
$$ 0 = A^3-A^2\operatorname{tr}{A} = 0-A^2\operatorname{tr}{A}, $$
so either $\operatorname{tr}{A}=0$ or $A^2=0$. But if $\operatorname{tr}{A}=0$, then (1) becomes $A^2=0$ anyway.