Attempt: Let $A,B\in\mathfrak{gl}(2,\mathbb{K})$ such that:
$$A = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}, \quad B = \begin{pmatrix}b_{11} & b_ {12} \\ b_{21} & b_{22}\end{pmatrix}$$
(This part is skippable). First, let's see their products:
$$AB = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix} = \begin{pmatrix}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12 }b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{pmatrix}$$
$$BA = \begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix} = \begin{pmatrix}b_{11}a_{11} + b_{12}a_{21} & b_{11}a_{12} + b_{12 }a_{22} \\ b_{21}a_{11} + b_{22}a_{21} & b_{21}a_{12} + b_{22}a_{22}\end{pmatrix}$$
Hence:
$$\begin{align*} AB - BA &= \begin{pmatrix}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21 }b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{pmatrix} - \begin{pmatrix}b_{11}a_{11 } + b_{12}a_{21} & b_{11}a_{12} + b_{12}a_{22} \\ b_{21}a_{11} + b_{22}a_{21} & b_{ 21}a_{12} + b_{22}a_{22}\end{pmatrix} \\[0.5em] &= \begin{pmatrix}a_{11}b_{11} + a_{12}b_{21} - b_{11}a_{11} - b_{12}a_{21} & a_{11}b_{12 } + a_{12}b_{22} - b_{11}a_{12} - b_{12}a_{22} \\ a_{21}b_{11} + a_{22}b_{21} - b_{ 21}a_{11} - b_{22}a_{21} & a_{21}b_{12} + a_{22}b_{22} - b_{21}a_{12} - b_{22}a_{22 }\end{pmatrix} \\[0.5em] &= \begin{pmatrix}a_{12}b_{21} - b_{12}a_{21} & a_{11}b_{12} + a_{12}b_{22} - b_{11}a_{12 } - b_{12}a_{22} \\ a_{21}b_{11} + a_{22}b_{21} - b_{21}a_{11} - b_{22}a_{21} & a_{ 21}b_{12} - b_{21}a_{12}\end{pmatrix} \\[0.5em] \end{align*}$$
Now, let's see what the center of $\mathfrak{gl}(2,\mathbb{K})$ looks like. By definition:
$$\begin{align*} \mathrm{Z}(\mathfrak{gl}(2,\mathbb{K})) &= \{A\in\mathfrak{gl}(2,\mathbb{K}) : [A,B] = 0 , \text{ for all }B\in\mathfrak{gl}(2,\mathbb{K})\} \\[0.5em] &= \{A\in\mathfrak{gl}(2,\mathbb{K}) : AB - BA = 0, \text{ for all }B\in\mathfrak{gl}(2,\mathbb{K} )\} \end{align*}$$
In other words, the center of $\mathfrak{gl}(2,\mathbb{K})$ is made up of the matrices $A\in\mathfrak{gl}(2,\mathbb{K})$ such that $ AB - BA = 0$, for all $B\in\mathfrak{gl}(2,\mathbb{K})$, that is:
$$\begin{align} a_{12}b_{21} - b_{12}a_{21} = 0 \qquad (1)\\[0.5em] a_{11}b_{12} + a_{12}b_{22} - b_{11}a_{12} - b_{12}a_{22} = 0 \qquad (2) \\[0.5em] a_{21}b_{11} + a_{22}b_{21} - b_{21}a_{11} - b_{22}a_{21} = 0 \qquad (3) \\[0.5em] a_{21}b_{12} - b_{21}a_{12} = 0 \qquad (4) \\[0.5em] \end{align}$$
At this point I don't know how to continue. Am I forgetting something? Please help.
