GAUSSIAN ELIMINATION WORKSHEET

Solve the following systems of linear equations by using the Gauss elimination method :

Problem 1 :

5x + 6y = 7

3x + 4y = 5

Problem 2 :

2x - 2y + 3z = 2

x + 2y - z = 3

3x - y + 2z = 1

Problem 3 :

2x + 4y + 6z = 22

3x + 8y + 5z = 27

-x + y + 2z = 2

Problem 4 :

4y + 2z = 1

2x + 3y + 5z = 0

3x + y + z = 11

Problem 5 :

3x + 6y - 9z = 15

2x + 4y - 6z = 10

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

1. Answer :

5x + 6y = 7

3x + 4y = 5

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

2. Answer :

2x - 2y + 3z = 2

x + 2y - z = 3

3x - y + 2z = 1

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 - 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

3. Answer :

2x + 4y + 6z = 22

3x + 8y + 5z = 27

-x + y + 2z = 2

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 = 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

4. Answer :

4y + 2z = 1

2x + 3y + 5z = 0

3x + y + z = 11

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

Add 5/4 to both sides.

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

Add 4 to both sides.

x = 4

Therefore the solution of the system is

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

5. Answer :

3x + 6y - 9z = 15

2x + 4y - 6z = 10

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

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.

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