USE GAUSSIAN ELIMINATION TO SOLVE THE SYSTEM OF EQUATIONS

Use Gaussian elimination to determine the solution set to the given system.

Problem 1 :

x + 2y + z = 1, 3x + 5y + z = 3 and 2x + 6y + 7z = 1

Solution :

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

-y - 2z = 0 --- (2)

z = -1 ---- (3)

Applying the value of z in (2), we get

-y - 2(-1) = 0

-y + 2 = 0

-y = -2

y = 2 

Applying the value of z = -1 and y = 2 in (1), we get

x + 2(2) - 1 = 1

x + 4 - 1 = 1

x + 3 = 1

x = 1 - 3

x = -2

So, the solution is (-2, 2, -1).

Problem 2 :

3x - y + 0z = 1, 2x + y + 5z = 4 and 7x - 5y - 8z = -3

Solution :

3x - y + 0z = 1 --- (1)

5y + 15z = 10 ---- (2)

0z = 0 --- (3)

z = 0

Applying the value of z in (2), we get

5y + 15(0) = 10

5y = 10

y = 10/5

y = 2

Applying the value of z = 0 and y = 2 in (1), we get

3x - 2 + 0z = 1

3x = 1 + 2

3x = 3

x = 3/3

x = 1

So, the solution is (1, 2, 0).

Problem 3 :

3x + 5y - z = 14, x + 2y + z = 3 and 2x + 5y + 6z = 2

Solution :

3x + 5y - z = 14 ---- (1)

y + 2z = -5 --- (2)

0z = 3 --- (3)

So, the system of equation has no solutions.

Problem 4 :

6x - 3y + 3z = 12, 2x - y + z = 4 and -4x + 2y - 2z = -8

Solution :

x - 0.5y + 0.5z = 2

So, the system of equation has infinitely many solutions.

Use Gaussian elimination to find the solution for the given system of equations.

Problem 5 :

2x + 5y  = 9, x + 2y - z = 3 and -3x - 4y + 7z = 1

Solution :

x + 2y - z = 3

y + 2z = 3

0z = 4

z = 0

So, the system of equation has no solutions.

Problem 6 :

The circle given by the equation x² + y² + ax + by + c = 0 passes through the points (-2, 0), (-1, 7) and (5, -1). Find a, b and c.

Solution :

Given, equation x² + y² + ax + by + c = 0 ---- (1)

(-2, 0) substitute the equation (1)

(-2)² + (0)² + (-2)a + 0y + c = 0

4 - 2a + c = 0

-2a + c = -4 --- (2)

(-1, 7) substitute the equation (1)

(-1)² + (7)² + (-1)a + 7b + c = 0

1 + 49 - a + + 7b + c = 0

50 - a + 7b + c = 0

-a + 7b + c = -50 --- (3)

(5, -1) substitute the equation (1)

(5)² + (-1)² + 5a + (-1)b + c = 0

25 + 1 + 5a - b + c = 0

26 + 5a  - b + c = 0

5a  - b + c = -26 --- (4)

Equation (2), (3) and (4) form the matrix form.

x - 0.5z = 2 ---- (1)

y + 0.1z = -6 --- (2)

3.6z = -42 --- (3)

z = -42/3.6

z = -12

Applying the value of z in (2), we get

y + 0.1(-12) = -6

y - 1.2 = -6

y = -6 + 1.2

y = -4.8

y = -5 

Applying the value of z = -12 and y = -5 in (1), we get

x - 0.5(-12) = 2

x + 6 = 2

x = 2 - 6

x  = -4 

So, the solutions of a,b and c is -4, -5, -12.