If $AB=BA$, prove that $ A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} $ 
Let $A$ a $2\times2$ matrix, if $AB=BA$ for every $B$ of the size $2\times2$, Prove that:
$$
A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}
$$
$a \in \mathbb{R}$

My attempt:
Let  $$
A=\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}
$$
$$B=\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}$$
And since $AB=BA$, then
$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$
So $b_2 c_1=b_1 c_2$
And 
$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$
$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$
But what can I do now ?
Thanks :)
 A: Note:
$$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \Rightarrow \\
\begin{cases}\require{cancel}\cancel{a_{11}b_{11}}+a_{12}b_{21}=\cancel{b_{11}a_{11}}+b_{12}a_{21}\\
a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\
a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\
a_{21}b_{12}+\cancel{a_{22}b_{22}}=b_{21}a_{12}+\cancel{b_{22}a_{22}}\end{cases}$$
From $(1)$, since $b_{12}$ and $b_{21}$ can be any number, in particular, $b_{12}=0$ and $b_{21}\ne 0$, we get: $a_{12}=0$. 
Similarly, for $b_{12}\ne 0$ and $b_{21}=0$, we get $a_{21}=0$.
From $(2)$, since $a_{12}=0$ and $b_{12}$ is an arbitrary number, we get $a_{11}b_{12}=b_{12}a_{22} \Rightarrow a_{11}=a_{22}$. 
A: Let $A=\pmatrix{a&b\\c&d}$. For $B=\pmatrix{1&0\\0&0}$ we have
$AB=\pmatrix{a&0\\c&0}$ and $BA=\pmatrix{a&b\\0&0}$. So $AB=BA$
implies $b=c=0$, that is $A=\pmatrix{a&0\\0&d}$. Now try, say $B=\pmatrix{0&1\\0&0}$.
