In mathematics, Gaussian elimination method is known as the row reduction algorithm for solving systems of linear equations. It consists of a sequence of operations performed on the corresponding matrix of coefficients.
To reduce the augmented matrix to row - echelon form you should follow the following steps :
Step 1 :
Locate the leftmost column that does not consist entirely of zeros.
Step 2 :
Interchange the top row with another row, if necessary, to bring a nonzero entry to the top of the column found in Step 1.
Step 3 :
If the entry that is now at the top of the column found in Step 1 is b, multiply the first row by 1/b in order to introduce a leading 1.
Step 4 :
Add suitable multiples of the top row to the rows below so that all entries below the leading 1 become zeros.
Step 5 :
Now cover the top row in the matrix and begin again with Step 1 applied to the submatrix that remains. Continue in this way until the entire matrix is in row - echelon form.
Solve the following systems of linear equations by Gaussian elimination method :
Example 1 :
2x - 2y + 3z = 2
x + 2y - z = 3
3x - y + 2z = 1
Solution :
The last matrix is in row - echelon form. The corresponding reduced system is :
x + 2y - z = 3 ----(1)
-6y + 5z = -4 ----(2)
-5z = -20 ----(3)
In (3), solve for z.
-5z = -20
Divide both sides by -5.
z = 4
Substitute z = 4 in (2).
-6y + 5(4) = -4
-6y + 20 = -4
Subtract 20 from both sides.
-6y = -24
Divide both sides by -6.
y = 4
Substitute y = 4 and z = 4 in (1).
x + 2(4) - 4 = 3
x + 8 - 4 = 3
x + 4 = 3
x = -1
Therefore the solution of the system is
x = -1, y = 4 and z = 4
Example 2 :
2x + 4y + 6z = 22
3x + 8y + 5z = 27
-x + y + 2z = 2
Solution :
The last matrix is in row - echelon form. The corresponding reduced system is :
x + 2y + 3z = 11 ----(1)
2y - 4z = -6 ----(2)
22z = 44 ----(3)
In (3), solve for z.
22z = 44
Divide both sides by 22.
z = 2
Substitute z = 2 in (2).
2y - 4(2) = -6
2y - 8 = -6
Add 8 to both sides.
2y = 2
Divide both sides by 2.
y = 1
Substitute y = 1 and z = 2 in (1).
x + 2(1) + 3(2) = 11
x + 2 + 6 = 11
x + 8 = 11
Subtract 8 from both sides.
x = 3
Therefore the solution of the system is
x = 3, y = 1 and z = 2
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