Can we show that all $2 \times 2$ matrices are sums of matrices with determinant 1? I came across a paper on the Sums of 2-by-2 Matrices with Determinant One. In the paper, which I have conveniently indicated here for reference, the author claims, but without proof, that a $2 \times 2$ is a sum of elements of the special linear group, $SL_2(\mathbb{F})$, whose elements, $U$, are also $2 \times 2$ matrices, such that $|U|=1$.
I was thinking of proving this by either technique. Let $A= \begin{bmatrix}a & b\\c & d\end{bmatrix}$, and $a, b, c, d \in \mathbb{F}$.
Technique 1. Consider the following 4 matrices with determinant 1: $M_1= \begin{bmatrix}e & 0\\0 & 1/e\end{bmatrix}$, $M_2 = \begin{bmatrix}0 & -f\\1/f & 0\end{bmatrix}$, $M_3 = \begin{bmatrix}1/h & 0\\0 & h\end{bmatrix}$, and $M_4 = \begin{bmatrix}0 & 1/g\\-g & 0\end{bmatrix}$. 
We show that that $\sum\limits_{i=1}^4 M_i = A$.
Thus, we have $e + 1/h = a$, $1/e + h = d$, $-f + 1/g = b$, $1/f - g = c$.
Technique 2. Consider $\{U_i\}_{i=1} ^\infty \in SL_2(\mathbb{F})$. Show that the sum of a countable number of $U_i$s is $A$. The problem I have here is that I don't know how to proceed from here.
I don't know if either of these will be considered correct, though.
I hope someone could help me out here. Thanks.
 A: It is sufficient to show that the matrices $\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right)$, and $\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right)$ can be written as the sum of matrices with determinant $1$.
Now, notice that
\begin{align}\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} \frac{a}{2} & 1 \\ -1 & 0 \end{matrix}\right) + \left(\begin{matrix} \frac{a}{2} & -1 \\ 1 & 0 \end{matrix}\right) \\
\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & \frac{b}{2} \\ 0 & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & \frac{b}{2} \\ 0 & -1 \end{matrix}\right) \\
\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & 0 \\ \frac{c}{2} & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & 0 \\ \frac{c}{2} & -1 \end{matrix}\right) \\
\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right) &=\left(\begin{matrix} 0 & 1 \\ -1 & \frac{d}{2} \end{matrix}\right) + \left(\begin{matrix} 0 & -1 \\ 1 & \frac{d}{2} \end{matrix}\right).
\end{align}
A: I was considering several cases when I realized that
$$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}= \begin{bmatrix}
a&1\\-1&0
\end{bmatrix}
+
 \begin{bmatrix}1&b\\0&1
\end{bmatrix}
+
 \begin{bmatrix}
-1&0\\c&-1
\end{bmatrix}
+\begin{bmatrix}0&-1\\1&d\end{bmatrix}.
$$
This avoids the problem of the first technique that requires dealing with the cases where some entries are zero.
As for the second technique,  I cannot imagine what is the idea.
A: The matrix decomposition into a matrices is not unique. For example, $$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}= \begin{bmatrix}
a&-1\\1&0
\end{bmatrix}
+
 \begin{bmatrix}-1&b\\0&-1
\end{bmatrix}
+
 \begin{bmatrix}
1&0\\c&1
\end{bmatrix}
+\begin{bmatrix}0&1\\-1&d\end{bmatrix}.
$$
is another such decomposition into matrices with a determinant of 1. 
Another possibility is to further decompose these four matrices individually. 
