Deleting a column from echelon form of a matrix is still in echelon form?

Question

This is a true or false question which demands a counter-example in case it's false. The question says:

If we delete any column of a matrix in echelon form, it will still remain in echleon form.

My attempt: Not necessarily. I took the example: $$ \begin{bmatrix} \fbox{1} & 2 & 3 & 0\\ 0 & \fbox{4} & 7 & 6\\ 0 & 0 & 0 & \fbox{9} \\ \end{bmatrix} $$ If we delete the first column, the matrix becomes: $$ \begin{bmatrix} 2 & 3 & 0\\ 4 & 7 & 6\\ 0 & 0 & 9\\ \end{bmatrix} $$ which doesn't appear to be in echleon form anymore according to me.

But my textbook says it's a true statement. I am not getting where am i going wrong.

EDIT

The definition of echleon form of a matrix as it appears in my textbook:

Answer 1

Pretty sure your textbook is just wrong. For a more simple example, consider deleting any column but the last from the identity matrix $I_n$: the resulting matrix will have a zero row that is not at the end, so will not be in echelon form, unless your textbook is using some non-standard definition of "echelon form".