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.
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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