This set of Numerical Methods Multiple Choice Questions & Answers (MCQs) focuses on “Jacobi’s Iteration Method”.

1. The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeroes along ________

a) Leading diagonal

b) Last column

c) Last row

d) Non-leading diagonal

Answer: a

Explanation: The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeroes along the leading diagonal because convergence can be achieved only through this way.

2. The Jacobi iteration converges, if A is strictly dominant.

a) True

b) False

Answer: a

Explanation: If A is matrix is said to be diagonally dominant if for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row, and for such matrices only Jacobi’s method converges to the accurate answer.

3. In Jacobi’s Method, the rate of convergence is quite ______ compared with other methods.

a) Slow

b) Fast

Answer: a

Explanation: In Jacobi’s Method, the rate of convergence is quite slow compared with other methods because here the selection of unknowns of an iteration is done using the results of the previous iteration only, whereas in other methods, selection of unknowns is done along with the generation of results in an iteration.

4. Which of the following is an assumption of Jacobi’s method?

a) The coefficient matrix has no zeros on its main diagonal

b) The rate of convergence is quite slow compared with other methods

c) Iteration involved in Jacobi’s method converges

d) The coefficient matrix has zeroes on its main diagonal

Answer: a

Explanation: This is because it is the method employed for solving a matrix such that for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row. This helps in converging the result and hence it is an assumption.

5. How many assumptions are there in Jacobi’s method?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are two assumptions in Jacobi’s method.

6. The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal.

a) True

b) False

Answer: a

Explanation: The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal because the desirable convergence of the answer can be achieved only for a matrix which is diagonally dominant and a matrix that has no zeros along its main diagonal can never be diagonally dominant.

7. Which of the following is another name for Jacobi’s method?

a) Displacement method

b) Simultaneous displacement method

c) Simultaneous method

d) Diagonal method

Answer: b

Explanation: Jacobi’s method is also called as simultaneous displacement method because for every iteration we perform, we use the results obtained in the subsequent steps and form new results.

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

20x + y – 2z = 17
3x + 20y – z = -18
2x – 3y + 20z = 25

a) x = 1, y = -1, z = 1

b) x = 2, y = 1, z = 0

c) x = 2, y = 1, z = 0

d) x = 1, y = 2, z = 1

Answer: a

Explanation: We write the equations in the form

x = (frac{1}{20}) (17 – y +2z)

y = (frac{1}{20}) (-18 -3x + z)

z = (frac{1}{20}) (25 -2x +3y)

We start from an approximation x = y = z = 0.

Substituting these in the right sides of the equations (i), (ii), (iii), we get

First iteration:

x = 0.85, y = -0.9, z = 1.25

Putting these values again in equations (i), (ii), (iii), we obtain,

x = [17 – (-0.9) + 2(1.25)] = 1.02

y = [-18 -3(0.85) + 1.25] = -0.965

z = [25 – 2(0.85) + 3(-0.9)] = 1.03

Substituting these values again in equations (i), (ii), (iii), we obtain,

Second iteration:

x = 1.00125, y = -1.0015, z = 1.00325

Proceeding in this way, we get,

Third iteration:

x = 1.0004, y = -1.000025, z = 0.9965

Fourth iteration

x = 0.999966, y = -1.000078, z = 0.999956

Fifth iteration

x = 1.0000, y = -0.999997, z = 0.999992

The values in the last iterations being practically the same, we can stop.

Hence the solution is

x = 1, y = -1, z = 1.

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

10x = y – x = 11.19
x + 10y + z = 28.08
-x + y + 10z = 35.61

correct to two decimal places.

a) x = 1.00, y = 2.95, z = 3.85

b) x = 1.96, y = 2.63, z = 3.99

c) x = 1.58, y = 2.70, z = 3.00

d) x = 1.23, y = 2.34, z = 3.45

Answer: d

Explanation: Rewriting the equations as,

x = (frac{1}{10}) (11.19 – y + z)

y = (frac{1}{10}) (28.08 – x – z)

z = (frac{1}{10}) (35.61 + x – y)

We start from an approximation, x = y = z = 0.

First iteration, x = 1.119, y = 2.808, z = 3.561

Second iteration,

x = (frac{1}{10}) (11.19 – 2.808 + 3.651) = 1.19

y = (frac{1}{10}) (28.08 – 1.119 – 3.561) = 2.34

z = (frac{1}{10}) (35.61 + 1.119 – 2.808) = 3.39

Third Iteration:

x =1.22, y =2.35, z =3.45

Fourth iteration:

x =1.23, y =2.34, z =3.45

Fifth iteration:

x =1.23, y =2.34, z =3.45

Hence, x = 1.23, y = 2.34, z = 3.45.

10. 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 = (frac{1}{10})(3 + 2b + c + d)

b = (frac{1}{10})(15 + 2z + c + d)

c = (frac{1}{10})(27 + a + b + 2d)

d = (frac{1}{10})(-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 = (frac{1}{10})[3 + 2(1.5) + 2.7 + (-0.9)] = 0.78

b = (frac{1}{10})[15 + 2(0.3) + 2.7 + (-0.9)] = 1.74

c = (frac{1}{10})[27 + 0.3 + 1.5 + 2(-0.9)] = 2.7

d = (frac{1}{10})[-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.