Solution to $AB=BA$ given $B$ a $2\times 2$ matrix and non singular? If $AB=BA$  given $B$ is non singular and a two dimensional matrix can we conclude that  $A$ is either a scalar multiple of $B^{-1}$  or a scalar multiple of identity $I$  or a linear combination of the two ? 
 A: No, we have the subgroup of $GL_2$ of matrices of the form
$\left(\begin{array}[cc]\\ 1&x\\0&1
\end{array}\right).$
This is a collection of commuting invertible matrices, so for any $A, B$ of this form, we have $AB=BA$, with $B$ nonsingular. But $A$ is not a scalar multiple of $B^{-1}$ nor a scalar matrix.
Edit: Given the motivation mentioned, I suggest working out what must be true with coordinates:
$B = \left(\begin{array}[cc]\\ a&b\\c&d
\end{array}\right), 
A = \left(\begin{array}[cc]\\ x&y\\z&w
\end{array}\right) $,
and work out what must be true of $A$. 
A: Let's assume $B$ is not scalar, otherwise any matrix $A$ will satisfy the identity. For a fixed $B$, consider the linear map
$$
f(A)=AB-BA
$$
from the vector space of $2\times 2$ matrices to itself. Since $B$ is not scalar, there is a matrix $A$ not commuting with $B$, so the rank of $f$ is at least $1$.
The matrices $I$, $B$ belong to the kernel of $f$ and are linearly independent. Let's compute $f$ on the standard matrices
$$
E_1=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad
E_2=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad
E_3=\begin{bmatrix}0&0\\1&0\end{bmatrix},\quad
E_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}
$$
We have
$$
f(E_1)=\begin{bmatrix}0&b\\-c&0\end{bmatrix},
f(E_2)=\begin{bmatrix}c&-a+d\\0&-c\end{bmatrix},
f(E_3)=\begin{bmatrix}-b&0\\a-d&b\end{bmatrix},
f(E_4)=\begin{bmatrix}0&-b\\c&0\end{bmatrix}
$$
so the rank of $f$ is the same as the rank of the matrix
$$
\begin{bmatrix}
0 & b & -c & 0 \\
c & -a+d & 0 & -c \\
-b & 0 & a-d & b \\
0 & -b & c & 0
\end{bmatrix}
$$
where $B=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$.
It is not difficult to show that the rank of the matrix is $2$ if either $b\ne0$ or $c\ne0$: just consider the top right or bottom left $2\times2$ minors. If $b=c=0$, then we know that $a\ne d$, so again the rank is $2$, by considering the central $2\times2$ minor.
Thus any matrix in the kernel is a linear combination of $I$ and $B$, or of $I$ and $B^{-1}$, because also $I$ and $B^{-1}$ are linearly independent and in the kernel.
