If $A^3 = O$, what condition do the entries of $A$ satisfy?

Question

$$A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$

When $A^3 = O$ (zero matrix) is satisfied, what condition do the four real numbers $a, b, c, d$ meet?

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I cubed $A$:

$A^3 = \begin{bmatrix} a(a^2+bc)+bc(a+d) & b(a^2+bc) + bd(a+d) \\ ac(a+d)+c(bc+d^2) & bc(a+d)+d(bc+d^2)\end{bmatrix}$

I don’t know how it is going to be.

Post script

Thank you for the replies, I might have found the answers:

$$ \begin{eqnarray} A^3_{11} * d - A^3_{12} * c = 0 \\ (ad-bc)(a^2+bc) = 0 \\ ad-bc = 0 \end{eqnarray} $$

But why $ a^2 + bc \neq 0$?

Answer 1

It seems the nontrivial solution is

$$\left( \begin{array}{cc} a & b \\ -\frac{a^2}{b} & -a \\ \end{array} \right)$$

Answer 2

Let $f(x)$ be the characteristic polynomial of $A.$ Since $A^3$ is zero, $f(x)$ divides $x^3.$ Therefore, $f(x)=x^2.$ Thus, $A$ has both trace and determinant zero. In other words, $d=-a$ and $bc=a^2.$ (If $a=0,$ observe that either $b$ or $c$ must zero.)