Why elementary row operation of Matrix doesn't change the rank of the Matrix ?? (Linear Algebra)

Why elementary row operation of Matrix doesn't change rank of the Matrix ?? (Mathematics)¶

"Elementary row operation" consists of three following activity.

• Row switching
  A row within the matrix can be switched wtih another row.
  $R_{i} \leftrightarrow R_{j}$

• Row multiplication
  Each element in a row can be multiplied by a non-zero constant.
  $kR_{i} \rightarrow R_{i}$ where $k \neq 0$

• Row addition
  Row can be replaced by the sum of that row and a multiple another row.
  $R_{i} + R_{j} \rightarrow R_{i}$ where $i \neq j$

Source :https://en.wikipedia.org/wiki/Rank_(linear_algebra)

Now I'd like to get into why "Elementary row operation of Matrix" doesn't change "rank" of that matrix.
First of all, I have to make it clear what the "rank" of matirx mean ??
One of the definition of "rank" of matrix $A$ is :

Dimension of the image of the linear transformation that is given by multiplication by A.

Therefore, here I'd like to talk about why "Elementary row operation " doesn't change the dimention of image that is given by multiplication by A.
In a word or two, "Elementary row operation" is correspond to multiplication by "regular matrix" :) Take a look at 3 activeties.

Let $A$ be matrix as bellow. $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{pmatrix} $$

1. Row switching
   It's equivalent to multiplication by identity matrix which interchange two rows as bellow.
   \begin{equation} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \end{equation}

   It goes without saying that determinant of regular matrix is 1. 　

2. Row multiplication
   It's equivalent to multiplication by identity matrix whose $ii$ element is multiplicated by constant $c$.
   \begin{equation} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ 3a_{21} & 3a_{22} & 3a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \end{equation}
   It goes without saying that determinant of regular matrix is -3.

3. Row addition
   It's euivalent to multiplication by identity matrix whose $ij$ element is constant $c$.
   \begin{equation} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} + 3a_{31} & a_{22} + 3a_{32} & a_{23} + 3a_{33} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \end{equation} It goes without saying that determinant of regular matrix is -1.

When think about mapping by regular matrix, since regular matrix has inverse matrix, mapping would be bijection. Therefore dimention of image which given by regular matrix is same as kernel image.
Consequently, given that a regular matrix is $P$, dimention of image given by $A$ is correspond to dimention of image given by $PA$ as following image.