# GAUSS ELIMINATION METHOD

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.

We can also use this method to estimate either of the following :

• The rank of a matrix.
• The determinant of a square matrix
• The inverse of an invertible matrix.

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 using the Gauss elimination method :

Example 1 :

5x + 6y = 7

3x + 4y = 5

Solution :

The system of linear equations has the following augmented matrix.

The last matrix is in row - echelon form. The corresponding reduced system is :

x + 6y/5 = 7/5 ----(1)

y = 2 ----(2)

Substitute y = 2 in (1).

x + 6(2)/5 = 7/5

x + 12/5 = 7/5

Subtract 12/5 from both sides.

x = 1

Therefore the solution of the system is

x = 1 and y = 2

Example 2 :

4y + 2z = 1

2x + 3y + 5z = 0

3x + y + z = 11

Solution :

The system of linear equations has the following augmented matrix.

The last matrix is in row - echelon form. The corresponding reduced system is :

x + 3y/2 + 5z/2 = 0 ----(1)

y + z/2 = 1/4 ----(2)

z = -5/2 ----(3)

Substitute z = -5/2 in (2).

y + (-5/2)/2 = 1/4

y - 5/4 = 1/4

y = 6/4

y = 3/2

Substitute y = 3/2 and z = -5/2 in (1).

x + 3(3/2)/2 + 5(-5/2)/2 = 0

x + 9/4 - 25/4 = 0

x - 16/4 = 0

x - 4 = 0

x = 4

Therefore the solution of the system is

x = 4, y = 3/2 and z = -5/2

Example 2 :

3x + 6y - 9z = 15

2x + 4y - 6z = 10

-2x - 3y + 4z = -6

Solution :

The system of linear equations has the following augmented matrix.

The last matrix is in row - echelon form. The corresponding reduced system is :

x + 2y - 3z = 5 ----(1)

y - 2z = 4 ----(2)

From (2),

y = 2z + 4

Let z = t.

y = 2t + 4

Substitute y = 2t + 4 and z = t.

x + 2(2t + 4) - 3t = 5

x + 4t + 8 - 3t = 5

x + t + 8 = 5

Subtract t and 8 from both sides.

x = -t - 3

The solution of the system is

x = -t - 3

y = 2t + 4

z = t

For different values of t, we will have different values for x, y and z.

Therefore the system has infinite number of solutions.

If t = 0,

x = -3

y = 4

z = 0

If t = 1,

x = -4

y = 6

z = 1

If t = -1,

x = -2

y = 2

z = -1

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