Question 863932
<pre>
The matrix 

{{{(matrix(3,4, 
 1, 2, 8, 0,
 0, 1, 3, 2,
 0, 0, 0, 0))}}}


is in row echelon form because:

(1) every row with any non-zeros has 1 as its as its
leftmost non-zero element (called its "leading 1".
(2) the leading 1's have no non-zero elements below them,
(3) the leading 1 on the 2nd row is farther to the
right than the leading row in the 1st row.
(4) the only all-zero row is at the bottom. 

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However, the matrix:

{{{(matrix(3,4, 
 1, green(2), 8, 0,
 0, red(1), 3, 2,
 0, 0, 0, 0))}}}

is NOT in REDUCED row-echelon form. 
That's because the leading (red) 1 in the 
second row has the non-zero (green) 2 ABOVE it.
To be in REDUCED row-echelon form, the matrix
must be in row echelon form, but also it must
have this additional property:

The leading 1's must have no non-zero elements ABOVE them.

To get it in reduced row-echelon form
we'd have to get a zero where the green 2
is.  So we'd need to multiply the second row
{{{(matrix(1,4, 0, red(1), 3, 2))}}} 
by -2, getting
{{{(matrix(1,4, 0, red(-2), -6, -4))}}}
and add it element by element to the first row:
{{{(matrix(1,4, 1+0, red(-2)+green(2), 8-6, 0-4))}}}
Getting:
{{{(matrix(1,4, 1,0, 2, -4))}}}
and replace the first row by that and get:

{{{(matrix(3,4, 
 1, 0, 2, -4,
 0, 1, 3, 2,
 0, 0, 0, 0))}}}

Now it's in reduced row-echelon form, because the
leading 1's have no elements above or below them,
and the 2nd row's leading 1 is further to the right
than the 1st row's leading 1.  Also the all-zero
row is at the bottom.

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The system of equations for which the  given  matrix 

{{{(matrix(3,4, 
 1, 2, 8, 0,
 0, 1, 3, 2,
 0, 0, 0, 0))}}}

is the augmented matrix is this system:

{{{system(1x+2y+8z=0, 0x+1y+3z=2,0x+0y+0z=0)}}}

Edwin</pre>