Every elementary matrix of the first kind is a commutator of unimodular matrices

Statement
Suppose $$k$$ is a field and $$n \ge 2$$ is a natural number. For $$i \ne j$$ in the set $$\{ 1,2,\dots,n\}$$, and $$\lambda \in k$$, let $$e_{ij}(\lambda)$$ be the matrix with $$\lambda$$ in the $$(ij)^{th}$$ position and $$0$$ elsewhere, and let $$E_{ij}(\lambda)$$ be the matrix with $$1$$s on the diagonal, $$\lambda$$ in the $$(ij)^{th}$$ position, and zeroes elsewhere. Matrices of the form $$E_{ij}(\lambda)$$ are termed elementary matrices.

Let $$SL_n(k)$$ denote the group of all matrices of determinant $$1$$ under multiplication. Clearly, $$E_{ij}(\lambda) \in SL_n(k)$$ for all $$i \ne j, \lambda \in k$$. The following are true:


 * If $$n \ge 3$$, every elementary matrix can be expressed as a commutator of two elementary matrices. In particular, it is the commutator of two elements of $$SL_n(k)$$.
 * If $$n = 2$$, and $$k$$ has more than three elements, every elementary matrix can be expressed as a commutator of two elements of $$SL_n(k)$$.

Facts used

 * 1) uses::Every elementary matrix is a commutator of elementary matrices: This statement is valid for $$n \ge 3$$.

The case $$n \ge 3$$
This follows from fact (1), and the observation that every elementary matrix is unimodular.

The case $$n = 2$$
If $$k$$ has more than three elements, there exists $$\mu \in k$$ such that $$\mu$$ is invertible and $$\mu^2 \ne 1$$.

We need to prove that for any $$\lambda$$, the element $$E_{12}(\lambda)$$ (and analogously, the element $$E_{21}(\lambda)$$) is expressible as a commutator. Indeed, set:

$$g = \begin{pmatrix} \mu & 0 \\ 0 & \mu^{-1} \end{pmatrix}, \qquad h = \begin{pmatrix} 1 & \lambda/(\mu^2 - 1) \\ 0 & 1 \end{pmatrix}$$

A computation shows that the commutator of $$g$$ and $$h$$ is $$E_{12}(\lambda)$$. A similar construction gives $$E_{21}(\lambda)$$ as a commutator.