Find All Matrices That Commute Find all $2\times 2$ matrices that commute with
$$\left( \begin{array}{cc}
12 & 9 \\
0 & -9 \end{array} \right)$$
My progress:
I know that we need to find all possible matrices A that AB = BA.
What I do is:
A := $$\left( \begin{array}{cc}
a & b \\
c & d \end{array} \right)$$
Then I multiple AB and set it equal to BA to get the following system of equations:
$12a = 12a+9c$
$9a-9b = 12b+9d$
$12c = -9c$
$9c-9d = -9d$
Doing so, I solve $c=0, a=1, d=1, b=-3/4$. However, given the answer choices, I feel like I did something wrong

Update: So I took another look. Since $c = 0, 12a=12a+9c$ tells us $a=a$. Then $9c-9d=-9d$ tells us $d = d$. That then tells us that the $9a-9b=12b+9d$ equation gives us $9a-9d=21b$, so $9(a-d)=21b$, so $9/21(a-d)=b$? So would the answer be NONE? Also how do I justify $\mathbb{R} $ vs $\mathbb{R}^2$?
 A: If$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\quad\text{and}\quad B=\begin{bmatrix}12&9\\0&-9\end{bmatrix},$$then$$AB-BA=\begin{bmatrix}-9 c & 9 a-21 b-9 d \\ 21 c & 9 c\end{bmatrix}.$$Therefore\begin{align}AB=BA&\iff\left\{\begin{array}{l}-9c=0\\9a-21b-9d=0\\21c=0\\9c=0\end{array}\right.\\&\iff\left\{\begin{array}{l}c=0\\b=\frac9{21}(a-d)\end{array}\right.\end{align}So, it seems that the correct answer is that none of the answers are correct, unless there was a typo. If that $12$ becomes a $10$, then the final option is correct.
A: Consider the linear map (commutator) $\phi: M_2(\mathbb{R})\to M_2(\mathbb{R})$ induced by your matrix $A$ sending $B$ to $AB-BA$.
Then you want to compute a basis of the kernel of the map.
Observe that $I, A\in \ker (\phi)$ and $I, A$ are linearly independent. Thus $\dim (\ker(\phi))\geq 2$.
However if it would be $3$, then the image has to be a line, but it’s not so because If you denote by $e_1,\cdots e_4$ the basis of the matrix space, then
$Ae_1-e_1A=12e_1-(12 e_1+9e_2)=-9e_2$
while
$Ae_2-e_2A=9e_2-9e_4-(-9e_4)=9e_2$
and
$Ae_3-e_3A=9e_1-9e_3-12e_3-9e_4=9e_1-21e_3-9e_4$
so $Ae_3-e_3A\not \in \langle e_2\rangle$
Thus the dimension of the image is not $1$ and so the dimension of the kernel has to be $2$, that implies
$\ker(\phi)=\langle I, A \rangle$
In any case it was clear that anyone of these were not correct because you can see that $A$ does not belongs to anyone of the possible $\Gamma$.
A: $$\left( \begin{array}{cc}
12 & 9 \\
0 & -9 \end{array} \right)\left( \begin{array}{cc}
a & b \\
c & d \end{array} \right)=\left( \begin{array}{cc}
a & b \\
c & d \end{array} \right)\left( \begin{array}{cc}
12 & 9 \\
0 & -9 \end{array} \right)$$
$$
\left( \begin{array}{cc}
12a+9c & 12b+9d \\
-9c & -9d \end{array} \right)=\left( \begin{array}{cc}
12a & 9a-9b \\
12c & -9c-9d \end{array} \right)
$$
$$12a+9c=12a \implies c=0$$
$$ 12b+9d=9a-9b \implies 21b+9d-9a=0 \hspace{5 pt} or \hspace{5 pt}7b+3d-3a=0 $$
Other equation leads to $c=0$
There are 3 variables to calculate and one equation. Let $d=k_1$  and $b=k_2 $ be arbitrary real numbers then
$$ a=\frac{(3k_1+7k_2)}{3}
$$
Now, set of all matrices which commute with the given matrix is given by
$$
C=\bigg\{ \left( \begin{array}{cc}
\frac{(3k_1+7k_2)}{3} & k_2 \\
0 & k_1 \end{array} \right) : k_1,k_2\in\mathbb{R} \bigg\}
$$
None of the options matches this answer hence None of the above is correct
