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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 first 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 entries in the column below the leading one be zero.
- Ignoring the top row, repeat steps 1 to 4 until there are no more leading ones.
- Finally, use each leading 1 to make all entries on the column above the zeros.
Example:
Solve for x, y, and z in:Solution:
2y - 3z = 2 2x + z = 3 x - y + 3z = 1 This is now in reduced row-echelon form, so we can read off the answer:
é
ê
ë0
2
12
0
-1-3
1
3ï
ï
ï2
3
1ù
ú
ûwrite the system as an augmented matrix
é
ê
ë1
2
0-1
0
23
1
-3ï
ï
ï1
3
2ù
ú
ûinterchange first and third row
(to make top left entry non-zero)
é
ê
ë1
0
0-1
2
23
-5
-3ï
ï
ï1
1
2ù
ú
ûAdd -2 times first row to second row
(to get 0 in first column of row 2)
é
ê
ë1
0
0-1
1
23
-5/2
-3ï
ï
ï1
1/2
2ù
ú
ûDivide second row by 2
(to get a leading 1 in row 2)
é
ê
ë1
0
0-1
1
03
-5/2
2ï
ï
ï1
1/2
1ù
ú
û
Add -2 times second row to third
(to get 0's in the second column)
é
ê
ë1
0
0-1
1
03
-5/2
1ï
ï
ï1
1/2
1/2ù
ú
ûDivide third row by 2
(to get a leading 1)
é
ê
ë1
0
0-1
1
00
0
1ï
ï
ï-1/2
7/4
1/2ù
ú
ûAdd -3 times third row to first row
Add 5/2 times third row to second row
(to get 0's in third column)
é
ê
ë1
0
00
1
00
0
1ï
ï
ï5/4
7/4
1/2ù
ú
ûAdd second row to first row x = 5/4 ,y = 7/4 , andz = 1/2 .Check that the answer satisfies the initial equations (in case we made arithmatic errors):
All of these check out, so our solution is correct.2y - 3z = 2(7/4) - 3(1/2) = 7/2 - 3/2 = 4/2 = 2
2x + z = 2(5/4) + 1/2 = 5/2 + 1/2 = 6/2 = 3
x - y + 3z = (5/4) - (7/4) + (1/2) = -(2/4) + 3/2 = -(1/2) + 3/2 = 2/2 = 1
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