@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

- Solve the system of equations by Jacobi’s iteration method.

10a - 2b - c - d = 3

- 2a + 10b - c - d = 15
- a - b + 10c - 2d = 27
- a - b - 2c + 10d = -9

a) a = 1, b = 2, c = 3, d = 0

b) a = 2, b = 1, c = 9, d = 5

c) a = 2, b = 2, c = 9, d = 0

d) a = 1, b = 1, c = 3, d = 5

Answer: a

Explanation: Rewriting the given equations as

a = 110(3 + 2b + c + d)

b = 110(15 + 2z + c + d)

c = 110(27 + a + b + 2d)

d = 110(-9 + a + b + 2d)

We start from an approximation a = b = c = d = 0.

First iteration: a = 0.3, b = 1.5, c = 2.7, d = -0.9

Second iteration:

a = 110[3 + 2(1.5) + 2.7 + (-0.9)] = 0.78

b = 110[15 + 2(0.3) + 2.7 + (-0.9)] = 1.74

c = 110[27 + 0.3 + 1.5 + 2(-0.9)] = 2.7

d = 110[-9 + 0.3 + 1.5 + 2(-0.9)] = -0.18

Proceeding in this way we get,

Third iteration, a = 0.9, b = 1.908, c = 2.916, d = -0.108

Fourth iteration, a = 0.9624, b = 1.9608, c = 2.9592, d = -0.036

Fifth iteration, a = 0.9845, b = 1.9848, c = 2.9851, d = -0.0158

Sixth iteration, a = 0.9939, b = 1.9938, c = 2.9938, d = -0.006

Seventh iteration, a = 0.9939, b = 1.9975, c = 2.9976, d = -0.0025

Eighth iteration, a = 0.999, b = 1.999, c = 2.999, d = -0.001

Ninth iteration, a = 0.9996, b = 1.9996, c = 2.9996, d = -0.004

Tenth iteration, a = 0.9998, b = 1.9998, c = 2.9998, d = -0.0001

Hence, a = 1, b = 2, c = 3, d = 0.