A matrix is said to be column reduced echelon form if
1. The leading entry of each non zero column is 1.
2. Each row containing the leading entry of some non zero column has all its other entries zero.
A matrix is said to be in column reduced echelon form if it satisfies the above two conditions, and also satisfy the following. 3. Each zero column lies to the right of a non zero column.
3. The leading non zero entry in any column is in the row that lies above the leading non zero entry in the next column.
The matrix
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 5 & 7 & 8 & 0 \\ 10 & 9 & 6 & 0 \\ 0 & 10 & 5 & 0 \end{array} \right)$
is to be reduced to column reduced echelon form
divide the column 1 by five.
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 7 & 8 & 0 \\ 2 & 9 & 6 & 0 \\ 0 & 10 & 5 & 0 \end{array} \right)$

Subtract seven times column 1 from column 2 and subtract eight times column 1 from column 3.
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 2 & -5 & -10 & 0 \\ 0 & 10 & 5 & 0 \end{array} \right)$

Divide column 2 by minus five
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 2 & 1 & -10 & 0 \\ 0 & -2 & 5 & 0 \end{array} \right)$

Subtract twice the second column from the first column and add ten times column 2 to column 3.
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 4 & -2 & -15 & 0 \end{array} \right)$

Divide column 3 by minus fifteen.
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 4 & -2 & 1 & 0 \end{array} \right)$

Subtract four times column 3 from column 1 and add two times column 3 to column 2.
$\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right)$

This is the column reduced echelon form of the matrix.
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