TITLE: What does it mean if $\det(A)$ equals $1$?
QUESTION [16 upvotes]: What does it mean if $\det(A)$ equals $1$? Does it mean that the identity matrix can be obtained from $A$ by only adding multiples of rows onto others?

REPLY [4 votes]: NOTE: My original answer was about a different question. (I have considered only adding integer multiples of some row.) I have completely changed it, hopefully the new answer is correct.

Other answers have already gave explanations about geometric meaning of $\det(A)=1$, so let me address the questions about row operations.
Let us start by checking what can be obtain from some given regular matrix $A$ using only one type of row operations. (Adding multiple of some row to another row.) 
We can exchange rows, with changing signs of one of them
If we want to exchange rows $\vec\alpha_1$ and $\vec\alpha_2$ we can proceed like this:
$$\begin{pmatrix}\vec\alpha_1\\\vec\alpha_2\end{pmatrix}\sim
\begin{pmatrix}\vec\alpha_1\\\vec\alpha_1+\vec\alpha_2\end{pmatrix}\sim
\begin{pmatrix}-\vec\alpha_2\\\vec\alpha_1+\vec\alpha_2\end{pmatrix}\sim
\begin{pmatrix}-\vec\alpha_2\\\vec\alpha_1\end{pmatrix}.$$
We can get from $A$ a diagonal matrix.
We proceed similarly as with Gaussian elimination, only we are not allowed to multiply rows, so we do not get $1$'s on diagonal, instead we obtain some non-zero elements.
We inductively repeat these two steps:

If in the $k$-th step the element $b_{kk}$ is non-zero we use it to get zeroes in all other rows in the $k$-th column.
If $b_{kk}=0$, at least one of elements $b_{k,k+1},\dots,b_{k,n}$ must be non-zero. (Otherwise we would have $\det(A)=0$.) We exchange rows in a such way that we get non-zero number on this position. Then we can proceed as before.

From the diagonal matrix, we can get identity matrix $I$.
We inductively repeat steps like these:
$$\begin{pmatrix}d_1&0\\0&d_2\end{pmatrix}\sim
\begin{pmatrix}d_1&0\\1&d_2\end{pmatrix}\sim
\begin{pmatrix}0&-d_1d_2\\1&d_2\end{pmatrix}\sim
\begin{pmatrix}0&-d_1d_2\\1&0\end{pmatrix}\sim
\begin{pmatrix}1&-d_1d_2\\1&0\end{pmatrix}\sim
\begin{pmatrix}1&0\\0&d_1d_2\end{pmatrix}$$
By repeating such step several times we can get to
$$\begin{pmatrix}
1 & 0 & \dots & 0 & 0 \\
0 & 1 & \dots & 0 & 0 \\
\ldots & \ldots & \ddots & \ldots & \ldots \\
0 & 0 & \dots & 1 & 0 \\
0 & 0 & \dots & 0 & d_1d_2\dots d_n
\end{pmatrix}=
\begin{pmatrix}
1 & 0 & \dots & 0 & 0 \\
0 & 1 & \dots & 0 & 0 \\
\ldots & \ldots & \ddots & \ldots & \ldots \\
0 & 0 & \dots & 1 & 0 \\
0 & 0 & \dots & 0 & 1
\end{pmatrix}=I_n
$$