Find all matrices that commute with given matrix Find all $2\times 2$ matrices that commute with
$$\left( \begin{array}{cc}
2 & 3 \\
1 & 4 \end{array} \right)$$
My progress:
I know that a square matrix commutes with itself, the identity matrix of that order, the null matrix of that order and any scalar matrix of that order.
The answer has been given as: 
$$\left( \begin{array}{cc}
m & 3n \\
n & m+2n \end{array} \right)$$
I don't understand how they're getting that form. Can someone please explain?
 A: You need $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)
\left(\begin{array}{cc}2&3\\1&4\end{array}\right)=
\left(\begin{array}{cc}2&3\\1&4\end{array}\right)\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Multiply out the matrices; that will give you four equations that connect $a,b,c$ and $d$.  Then solve those equations.
A: If you write down the unknown matrix as 
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$$
Then you want
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \cdot
\begin{bmatrix}
2 & 3 \\
1 & 4
\end{bmatrix}
=
\begin{bmatrix}
2 & 3 \\
1 & 4
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$$
Write out the left and right sides as matrices with entries like 2a + 1b, etc. Set them equal. You get four equations in 4 unknowns. Solve, and you get the answer above. 
A: Let's call your matrix $$A = \left( \begin{array}{cc}
2 & 3 \\
1 & 4 \end{array} \right)$$
We want a matrix $X_{2\times 2} = \begin{pmatrix} a & b\\ c&d\end{pmatrix}$ such that $AX = XA$.
$$AX = \begin{pmatrix} 2a + 3c & 2b+3d\\ a + 4c&b+4d\end{pmatrix}$$
$$XA = \begin{pmatrix} 2a + b&3a + 4b\\2c+d & 3c+4d\end{pmatrix}$$
Now you have a system of equations in four variables:
$$2a + 3c = 2a + b \implies b = 3c$$
$$2b+3d = 3a + 4b$$
$$a + 4c = 2c + d$$
$$b+4d= 3c+4d$$
Solve the system of equations. (Note, if you do Gaussian Elimination, you'll have two of the four rows all zero.)
