Math 15H (Notes)

Gauss-Jordon Elimination Notes:

The procedure for Guass-Jordan 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:

\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 non-zero)

\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)

Now we can read off the answer: x = 5/4, y = 7/4, and z = 1/2.
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.

Math 15H (Fall 2004) web pages
Created: 04 Sep 2004
Last modified: 13 Nov 2004 09:49:34
Comments to: dpvc@union.edu


			Matrix Inverse Notes
			Selected Course Notes


\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 non-zero)

\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)