The procedure for GuassJordan Elimination is as follows:
 Find the leftmost column that is not all zeros and swap its row with the top row.
 Make the leading entry in the top row a "1". (If the top entry is a, then multiply the top row by 1/a).
 Use the top row to make all the other entries in the column containing the leading one into zeros.
 Ignoring the top row, repeat steps 1 to 4 until there are no more leading ones.
Example:
Solve for x, y, and z in:\begin{eqnarray} 2y  3z &= 2\cr 2x + z &= 3\cr x  y + 3z &= 1\cr \end{eqnarray}Solution:Now we can read off the answer: x = 5/4, y = 7/4, and z = 1/2.
\Matrix{ 0& 2& 3& & 2\cr 2& 0& 1& \& 3\cr 1& 1& 3& & 1\cr }write the system as an augmented matrix \Matrix{ 1& 1& 3& & 1\cr 2& 0& 1& \& 3\cr 0& 2& 3& & 2\cr }interchange first and third rows
(to make top left entry nonzero) \Matrix{ 1& 1& 3& & 1\cr 0& 2& 5& \& 1\cr 0& 2& 3& & 2\cr }Add 2 times first row to second row
(to get 0 in first column of row 2) \Matrix{ 1& 1& 3& & 1\cr 0& 1& 5/2& \& 1/2\cr 0& 2& 3& & 2\cr }Divide second row by 2
(to get a leading 1 in row 2) \Matrix{ 1& 0& 1/2& & 3/2\cr 0& 1& 5/2& \& 1/2\cr 0& 0& 2& & 1\cr }Add second row to first row
Add 2 times second row to third
(to get 0's in the second column) \Matrix{ 1& 0& 1/2& & 3/2\cr 0& 1& 5/2& \& 1/2\cr 0& 0& 1& & 1/2\cr }Divide third row by 2
(to get a leading 1) \Matrix{ 1& 0& 0& & 5/4\cr 0& 1& 0& \& 7/4\cr 0& 0& 1& & 1/2\cr }Add 1/2 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)Check that the answer satisfies the initial equations (in case we made arithmatic errors):
\begin{array}{l} 2y  3z = 2(7/4)  3(1/2) = 7/2  3/2 = 4/2 = 2\cr 2x + z = 2(5/4) + 1/2 = 5/2 + 1/2 = 6/2 = 3\cr x  y + 3z = (5/4)  (7/4) + 3(1/2) = (2/4) + 3/2 = (1/2) + 3/2 = 2/2 = 1\cr \end{array}All of these check out, so our solution is correct.

