Can I exchange column and then use the row transformation when converting a matrix into a row echelon form?

Question

matrix-rank: Please help check if the below transformations are allowed for finding the rank.

Answer 1

Mostly no. There are a bunch of theorems about row-equivalent matrices (i.e. matrices that can be transformed to each other with a sequence of row operations) and swapping columns will not make you row-equivalent. The resulting matrix cannot be called a row-echelon form of the original matrix.

That said, when finding rank, column operations will also preserve rank, like row operations (as column rank and row rank are always equal). So, the row-echelon form matrix you find, though it doesn't correspond to the original matrix, it will still have the same rank, and the conclusion that the rank is $2$ is valid.

You can also potentially swap columns when using matrices to solve systems of equations. Each column corresponds to a variable in your system of equations. If you swap columns, this corresponds to swapping which variable corresponds to which column. So long as you keep track of this until the end of the process, you should end up with the correct solution.

Answer 2

When you try to find the rank of a matrix, you are allowed to use row and column operations simultaneously.

Just a small correction, what you are looking for is very different from finding the row echelon form of the matrix.

In fact, there is a very important result in linear algebra that states that : given a $ m\times n$ matrix $A$ with rank $r$, then $r\leq m$, $ r\leq n$ and by means of a finite number of elementary row and column operations, $A$ can be transformed into the matrix $$\begin{pmatrix} I_r & O_1 \\ O_2 & O_3 \end{pmatrix}$$ where $I_r$ is the $r\times r$ identity matrix and $O_1$, $O_2$ and $O_3$ are zero matrices.

To make it clearer, in your specific example, after applying row and column operations in $ A $ we end up in $$\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}$$ (again, which is not the rref of $A$). So, $\operatorname{rank}(A)=2$.

Answer 3

The rank equals the number of pivot columns, also known as the number of basic variables. So, the process is to row-reduce the matrix and count the number of pivots, as you did. So, your approach is valid. Since there are two pivots, the rank of the matrix is 2.