Driving the state of a discrete system to zero in one step I have the following system of difference equations:
$\textbf{x}(k+1) = A \textbf{x}(k) + \textbf{b} u(k)$
where:
$A = \begin{bmatrix}
1 & 2 \\
3 & \alpha
\end{bmatrix}
$ and, 
$\mathbf{b} = \begin{bmatrix}
1\\
1
\end{bmatrix}$
.. and am asked to find the values of $\alpha$ for which the state of the system can be driven to zero in just one step if $x(0) = \begin{bmatrix}
1\\
-2
\end{bmatrix}$
I have tried substituting $x(0)$ into $x(k)$ and $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ into $x(k+1)$ to give me two equations. Whilst the first equation results in a contradiction, i.e. the $LHS \neq RHS$, the second equation gives me $\alpha = 0.5$. Is my method for determining $\alpha$ correct?

You have a discrete system:
$x(k+1) = A x(k) + b u(k)$
where
$A = \begin{bmatrix}
1 & 2 \\
3 & \alpha
\end{bmatrix}
$ and
$b = \begin{bmatrix}
1\\
1
\end{bmatrix}$
and you want to find $u(k)$ such that $x(k + 1) = 0$ from the initial condition 
$$
x(0) = \begin{bmatrix}1 \\ -2 \end{bmatrix}
$$
So just insert everything:
$$
\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix}
1 & 2 \\
3 & \alpha
\end{bmatrix} \begin{bmatrix}1 \\ -2 \end{bmatrix} + \begin{bmatrix}1 \\ 1\end{bmatrix} u(k)
$$
Multiply these out you get:
$$
\begin{align}
u(k) - 3 &= 0 \\
u(k) - 2 \alpha + 3 &= 0
\end{align}
$$
So from the first equation we see that $u(k) = 3$. Insert that into the second equation you get:
$$
6 - 2\alpha = 0
$$
So you can see that your problem can be solved if and only if $\alpha = 3$.
