How to extend the matrix with determinant 1 to keep it Lets consider 2x2 integer matrix with determinant equal 1:
$$\left(
\begin{array}{cc}
 a & b \\
 c & d \\
\end{array}
\right)$$
I am working on the following:
How to extend this to 3x3 matrix in order to get another matrix with determinant 1:
$$\left(
\begin{array}{ccc}
 a & b & e \\
 c & d & f \\
 i & h & g \\
\end{array}
\right)$$
And also is there any $a,b,c,d$ for which this extension is unique.
I even have no idea how to start solving this. 
I have discovered the following so far on the web, but not sure how to use this:
Integer matrices with determinant equal to $1$
https://mathoverflow.net/questions/24131/is-the-semigroup-of-nonnegative-integer-matrices-with-determinant-1-finitely-gen
EDITED:
Actually I am looking for general algorithm, how to construct all 3x3 matricies from 2x2 matrix with determinant 1.
EDITED 2:
Some samples of such matricies:
$$\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 -1 & 0 & 1 \\
 -1 & 0 & 2 \\
\end{array}
\right)
$$
$$\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 1 & 2 & 3 \\
 2 & 5 & 9 \\
\end{array}
\right)$$
$$\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 -6 & -5 & -4 \\
 9 & 5 & 2 \\
\end{array}
\right)
$$
 A: You can just set $g = 1$ and $e,f,i,h = 0$:
$$1 = \begin{vmatrix}
 a & b \\
 c & d \\
\end{vmatrix} = \begin{vmatrix}
 a & b & 0\\
 c & d & 0\\
 0 & 0 & 1
\end{vmatrix}$$
Furthermore, the extension is never unique since:
$$1 = \begin{vmatrix}
 a & b \\
 c & d \\
\end{vmatrix} = \begin{vmatrix}
 a & b & 0\\
 c & d & 0\\
 0 & 0 & 1
\end{vmatrix} = \begin{vmatrix}
 a & b & 0\\
 c & d & 0\\
 a & b & 1
\end{vmatrix}$$
Note that $a$ and $b$ cannot both be $0$, so the two extensions are indeed different.
A: $SL_2 \mathbb Z$
$$
\left(
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right)
$$
homomorphic image
$SL_3 \mathbb Z$
$$
\left(
\begin{array}{ccc}
\alpha^2 & 2 \alpha \beta &  \beta ^2\\
\alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta\\
\gamma^2    & 2 \gamma \delta  & \delta^2
\end{array}
\right) 
$$
If you take the transpose of the 3 by 3, you get an anti-homomorphism. 
A: Hint:
Consider a block-diagonal matrix, for which the first diagonal block is $\;\begin{matrix}a&b\\c&d\end{matrix}$.
