How do I find the centralizer for this matrix? 
Let $B=\begin{pmatrix}
2 & 2 \\
 0& 1 \\
\end{pmatrix} \in GL_2(\mathbb{R})$. Determine the centralizer for $B$.

Here's what I've done so far:
Let $D=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \in GL_2(\mathbb{R})$.
$\begin{align} 
BD&=DB\\
\begin{pmatrix}
2 & 2 \\
 0& 1 \\
\end{pmatrix} \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} &= \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \begin{pmatrix}
2 & 2 \\
 0& 1 \\
\end{pmatrix}\\
\begin{pmatrix}
2(a+c) & 2(b+d) \\
 c& d \\
\end{pmatrix}&= \begin{pmatrix}
2a & 2a+b \\
 2c& 2c+d \\
\end{pmatrix}
\end{align} $
From the above, we have:
$\begin{aligned}
2a+2c&=2a\\
2b+2d&=2a+b\\
c&=2c\\
d&=2c+d
\end{aligned}
$
From which we derive:
$\begin{aligned}
c&=0\\
b+2d-2a&=0\\
d&=d \in \mathbb{R} \backslash \{0\}
\end{aligned}$
As you can see, I've managed to determine $c$ and $d$, but I can't seem to figure out $a$ and $b$. What am I missing?

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\R}{\mathbb{R}}$Allow me to sketch a slightly more conceptual approach.
$B$ has clearly the eigenvalues $2$ and $1$, hence it can be diagonalised, and one sees immediately that
$$
T^{-1} B T = \begin{bmatrix}2 & 0\\0 & 1\end{bmatrix},
\quad
\text{for }
T
=
\begin{bmatrix}1 & -2\\0 & 1\end{bmatrix}.
$$
Now the centraliser of $T^{-1} B T$ is immediately seen to be
$$
\mathbf{C} 
=
\Set{
\begin{bmatrix}a & 0\\0 & d\end{bmatrix}
:
a, d \in \R \setminus \Set{0}}.
$$
Hence the centraliser of $B$ is
$$
T \mathbf{C} 
 T^{-1}
=
\Set{
\begin{bmatrix}a & 2 a - 2 d\\0 & d\end{bmatrix}
:
a, d \in \R \setminus \Set{0}}.
$$
