= Solution
Put $u(t)=\begin{pmatrix}1&t\\0&1\end{pmatrix}$ and $l(s)=\begin{pmatrix}1&0\\s&1\end{pmatrix}$. These <elementary matrices> belong to the <special linear group>. Let $g=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ have determinant one. If $a\ne0$, elementary row elimination gives
$$
l(-c/a)g=\begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}
=\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}u(b/a).
$$
The remaining diagonal matrix is also a product of the given generators. Indeed, for $a\ne0$, direct multiplication gives
$$
w(a):=u(a)l(-a^{-1})u(a)=\begin{pmatrix}0&a\\-a^{-1}&0\end{pmatrix},\qquad
w(a)w(-1)=\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}.
$$
Hence $g=l(c/a)w(a)w(-1)u(b/a)$. If $a=0$, its <determinant> forces $c\ne0$, and $u(1)g$ has upper-left entry $c$. The established case applies to it, and $g=u(-1)(u(1)g)$ is again generated. No division by two was used, so the proof works over every <field>. \b[The upper and lower elementary unipotent matrices generate $SL_2(K)$.]
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