12:59 PM
room topic changed to Linear algebra: For any discussion concerning linear algebra [linear-algebra]
For instructions how to render MathJax(TeX) in chat see this post on meta.
I guess this might serve as a room of linear algebra-related topics, is such need should arise.
However, I have created it in connection with this question:
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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?

The OP asks whether every matrix such that $\det(A)=1$ can be transformed to identity matrix using only adding multiple of some row to another row.
It suffices to get diagonal matrix.
I think that if we are able to get a diagonal matrix, then the rest is easy.
We can inductively use this:
$\begin{pmatrix}d_1&0\\0&d_2\end{pmatrix}\sim$
$\begin{pmatrix}d_1&0\\1&d_2\end{pmatrix}\sim$
$\begin{pmatrix}0&-d_1d_2\\1&d_2\end{pmatrix}\sim$
$\begin{pmatrix}0&-d_1d_2\\1&0\end{pmatrix}\sim$
$\begin{pmatrix}1&-d_1d_2\\1&0\end{pmatrix}\sim$
$\begin{pmatrix}1&0\\0&d_1d_2\end{pmatrix}$
How to get diagonal matrix.
Since $\det(A)=1$, we know that $A$ is regular and that there is a matrix $B$ such that $BA=I$.
The equality $BA=I$ says that we are able to obtain $I$ by linear combinations of rows from $A$.
For example the linear combination with coefficients $b_{11},b_{12},...,b_{1n}$ yields the row $(1,0,....,0)$.
However, we are only allowed to replace the first row by a linear combination with coefficients $1,a_2,a_3,\dots,a_n$. Similarly for the second rows we are only allowed to use $b_1,1,b_3,\dots,b_n$.
This would correspond to multiplying from the left by a matrix which has all diagonal entries equal to $1$.
Well, if $b_{ii}\ne0$ where $B=A^{-1}$, then we can take the matrix $B'$ such that $i$-th row is $1/b_{ii}$-multiple of the $i$-th row of $B$.
For this matrix we have $B'A=D$, where $D$ is a diagonal matrix.
So in such case we can obtain a diagonal matrix using only allowed operation.
But what if some $b_{ii}=0$?
The algorithm I have suggested above wound not work for $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

1:48 PM
One more problem.
Even if we solve somehow the problem with diagonal elements, the solution I have suggested above still has gaps.
I argued that $BA$ can be obtained from $A$ using linear combinations where coefficients are determined by the rows of the matrix $A$.
But this is not the same as doing the row operations, where only a multiple of one row is added to another.
Here I am using the rows of the original matrix $A$. After the first step the rows are changed.

2:07 PM
ok, I think I solved this.
I hope I haven't made a mistake there.
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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 ...