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


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".
