Find a matrix so that $A^2$ not equal to 0 but $A^3$ is [Strang P78 2.4.23] 
(a) Find a nonzero matrix $A$ for which $A^2 = 0$.
  (b) Find a matrix that has $A^2 \neq 0$ but $A^3 = 0$.

Solution for (a): Let $A := \text{column $\times$ row} = \mathbf{cr^T} \neq \mathbf{0}$ where $\mathbf{c,r}$ are column vectors.
I'd like: $A^2 = \mathbf{c\color{green}{r^Tc}r^T} = \mathbf{0}. \mathbf{\color{green}{r^Tc = 0}}$ would imply this.
Thus, for want of a counterexample, choose $\mathbf{r} = (0,k)$ and $\mathbf{c} = (k,0)^T. \quad \blacksquare$
(b) Since I chose $\mathbf{c, r}$ such that  $\mathbf{\color{green}{r^Tc = 0}}$, thus $A^3 = \mathbf{c\color{green}{r^Tc}\color{green}{r^Tc}r^T} = 0.$  
$\Large{1.}$  Could someone please reveal and expound on the intermediate steps and thoughts towards devising a (counterexample) for $A^3$? Please don't answer with just a counterexample.
$\Large{2.}$ How would one foreknow/previse to define $A := \text{column $\times$ row}$?
 A: You want there to be a vector $v$ with $A^2\cdot v\neq0$ but $A^3\cdot v=0$. You can easily check that then $v,A\cdot v,A^2\cdot v$ must be a free (i.e., linearly independent) family: if there were a nontrivial linear dependence, take one with a minimal number of nonzero coefficients (at least two since the individual vectors cannot be zero); now apply $A$ sufficiently often to kill off the final nonzero coefficient, leaving exactly one less nonzero coefficient, which gives a contradiction.
For an easy example, take the (sub)space generated by $v,A\cdot v,A^2\cdot v$. The matrix of $A$ on this basis will be
$$
  A'=\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}
$$
which provides your example.
My argument shows that you need $A$ to have rank at least$~2$ (as $A\cdot v$ and $A^2\cdot v$ are linearly independent elements of the image of $A$); hence taking for $A$ a product $cr^T$ of a $n\times 1$ and a $1\times n$ matrix will never work.
A: You find yourself in a house with 3 rooms and no doors.
If you fall asleep in the first room, you disappear.
If you fall asleep in the second room, you have a 50% chance of waking up in the first room, otherwise you disappear.
If you fall asleep in the third room, you have a 33% chance of waking up in the first room, 33% chance of waking up in the second room, otherwise you disappear.
On arrival you could be in any room.  On the first morning you can only wake up in the first or second room.  On the second morning you can only wake up in the first room.  On the third morning there is nowhere you can be.
You describe your situation with a transition matrix.
$$ M = \begin{bmatrix} 0 & 0 & 0 \\ \frac 12 & 0 & 0 \\ \frac 13 & \frac 13 & 0 \end{bmatrix} $$
$$ M^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \frac 16 & 0 & 0 \end{bmatrix} $$
$$ M^3 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
A: A can be the matrix such that, multiplying any vector (x, y, z) by A, changes the vector by "pushing" every number in the vector down one position, and putting 0 at the top. That is, A(x, y, z)=(0, x, y).
So, multiplying any vector by A^2 will not always result in 0, but multiplying any vector by A^3 will always result in 0. This meets the requirements in the question.
To find the entries of this A, multiply A by each vector in the standard basis, and these results are the columns of A. So we get 
\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}
