Find all the $2 \times 2$ matrices $A$ satisfying $A^2 = 0$ 
Find all the $2 \times 2$ matrices $A$ satisfying $A^2 = 0$.

So far I have this:

But I don't know to proceed.
 A: Hint Two of the constituent equations of the matrix equality you've written are
$$a^2 + b c = 0 \qquad \textrm{and} \qquad b c + d^2 = 0.$$
Subtracting these leaves
$$a^2 - d^2 = 0 .$$

 Additional hint But $a^2 - d^2 = (a + d) (a - d)$, so either $d = a$ or $d = -a$. Work out the consequences of each of these equations for $b, c$.

Alternatively, here's a more abstract approach (which foreshadows some important topics in linear algebra):
If every vector $\bf v \in \Bbb R^2$ satisfies $A {\bf v} = 0$, then $A = 0$.
Otherwise, that is, if there is some vector $\bf v$ such that $A {\bf v} \neq 0$, then $A(A {\bf v}) = A^2 {\bf v} = {\bf 0}$, and so the matrix representation of the linear transformation ${\bf v} \mapsto A {\bf v}$ with respect to the basis $(A{\bf v}, {\bf v})$ is
$$\pmatrix{0&1\\0&0}.$$ Then, the change of basis formula tells us that $A$ must have the form
$$P^{-1}\pmatrix{0&1\\0&0}P .$$
for some invertible matrix $P$.

 Since scaling $P := \pmatrix{p&q\\r&s}$ doesn't change this product, we may as well scale the matrix so that $\det P = p s - q r = \pm 1$. Substituting, we get that the matrices satisfying $A^2 = 0$ are exactly those of the form $$\pm\pmatrix{rs&s^2\\-r^2&-rs} .$$

