This set of Numerical Methods Multiple Choice Questions & Answers (MCQs) focuses on “Gauss Seidel Method”.

1. Which of the following systems of linear equations has a strictly diagonally dominant coefficient matrix?

a)

3x – y = -4 2x + 5y = 2

b)

2x + y = 1 x - 7y = 4

c)

3x + 5y = 2 x + y = -3

d)

4x = 2y – z - 1 x + z = -4 3x – 5y + z = 3View Answer

Explanation: The coefficient matrix

A = \(\begin{bmatrix}3&-1\\2&5\end{bmatrix}\)

is strictly diagonally dominant because |3| > |-1| and |5| > |2|.

2. Gauss Seidal method is also termed as a method of _______

a) Successive displacement

b) Eliminations

c) False positions

d) Iterations

View Answer

Explanation: Gauss Seidal method is also termed as a method of successive displacement because with the iterations, we keep on displacing the values we select for the unknowns which are used in the subsequent steps.

3. Gauss seidal method is similar to which of the following methods?

a) Iteration method

b) Newton Raphson method

c) Jacobi’s method

d) Regula-Falsi method

View Answer

Explanation: The procedure involved in Gauss seidal and Jacobi is almost same. The only difference is that the while selecting the values for the unknowns, they use different methodology.

4. What is the main difference between Jacobi’s and Gauss-seidal?

a) Computations in Jacobi’s can be done in parallel but not in Gauss-seidal

b) Convergence in Jacobi’s method is faster

c) Gauss seidal cannot solve the system of linear equations in three variables whereas Jacobi cannot

d) Deviation from the correct answer is more in gauss seidal

View Answer

Explanation: Computations in Jacobi’s can be done in parallel but not in Gauss-seidal because in Jacobi’s method, the entire set of values obtained during the previous iteration is used as it is in the next one, whereas in Gauss-seidal method, as we keep on getting the individual values of the variable, we use them in the subsequent steps.

5. While solving by Gauss Seidal method, which of the following is the first Iterative solution system; x – 2y = 1 and x + 4y = 4?

a) (1, 0.75)

b) (0.25,1)

c) (0,0)

d) (1,0.65)

View Answer

Explanation: Here,

x – 2y = 1

x + 4y = 4

For first iteration we put n = 0 in the following equations,

x

_{n+1}= 1 – 2y

_{n}

y

_{n+1}= (1/4) (4 – x

_{n+1})

We get,

x

_{1}= 1 – 2(0) = 1

y

_{1}= (1/4) (4 – 1) = 0.75.

6. The Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric________ definite matrices.

a) Positive

b) Negative

c) Zero

d) Equal

View Answer

Explanation: Gauss-Seidel method is applicable to strictly diagonally dominant or symmetric positive definite matrices because only in this case convergence is possible.

7. Gauss seidal requires less number of iterations than Jacobi’s method.

a) True

b) False

View Answer

Explanation: Gauss-seidal requires less number of iterations than Jacobi’s method because it achieves greater accuracy faster than Jacobi’s method. This is the modification made to Jacobi’s method, which is now called as Gauss-seidal method.

8. Which of the following method is employed for solving the system of linear equations?

a) Runge Kutta

b) Newton Raphson

c) Gauss Seidal

d) Simpson’s Rule

View Answer

Explanation: Runge Kutta method is used to solve differential equations. Newton-Raphson method is used to find the root of a polynomial. Simpson’s rule is used to solve integration problems. Gauss-seidal is used for solving system of linear equations.

9. What is the limitation of Gauss-seidal method?

a) It cannot be used for the matrices with non-zero diagonal elements

b) It is more complex than Jacobi’s method

c) It doesn’t guarantees convergence for each and every matrix

d) It is an iterative technique

View Answer

Explanation: It does not guarantee convergence for each and every matrix. Convergence is only possible if the matrix is either diagonally dominant, positive definite or symmetric.

10. Solve the following equations by Gauss seidal 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 = 5, c = 7, d = -1

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

c) a = 2, b = 5, c = 3, d = 5

d) a = 2, b = 2, c = 8, d = -3

View Answer

Explanation:

a = \(\frac{1}{10}\)(3 + 2b + c + d) …………….(i)

b = \(\frac{1}{10}\)(15 + 2a + c + d) ……………(ii)

c = \(\frac{1}{10}\)(27 + a + b +d) ……………..(iii)

d = \(\frac{1}{10}\)(-9 + a + b + 2c) ……………(iv)

First iteration

Putting b = 0, c = 0, d = 0 in (i), we get a = 0.3

Putting a = 0.3, c = 0, d = 0 in (ii), we obtain b = 1.56

Putting a = 0.3, b = 1.56, d = 0 in (iii), we obtain c = 2.886

Putting a = 0.3, b = 1.56, c = 2.886 in (iv), we get d = -0.1368

Second iteration

Putting b = 1.56, c = 2.886, d = -0.1368 in (i), we obtain a = 0.8869

Putting a = 0.8869, c = 2.886, d = -0.1368 in (ii), we obtain b = 1.9523

Putting a = 0.8869, b = 1.9523, d = -0.1368 in (iii), we have c = 2.9566

Putting a = 0.8869, b = 1.9523, c = 2.9566 in (iv) we get d = -0.0248

Third iteration

Putting b = 1.9523, c = 2.9566, d = -0.0248 in (i), we obtain a = 0.986

Putting a = 0.9836, c = 2.9566, d = -0.0248 in (ii), we obtain b = 1.9899

Putting a = 0.9836, b = 1.9899, d = -0.0248 i (iii), we get d = 2.9924

Putting a = 0.9836, b = 1.9899, c = 2.9924 in (iv), we get d = -0.0042

Fourth iteration proceeding as above

a = 0.9968 b = 1.9982, c = 2.9987, d = -0.0008

Fifth iteration is a = 0.9994, b = 1.9997, c = 2.9997, d = -0.0001

Sixth iteration is a = 0.9999, b = 1.9999, c = 2.9999, d = -0.0001

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

11. Solve the following equation by Gauss Seidal Method up to 2 iterations and find the value of z.

27x + 6y - z = 85 6x + 15y + 2z = 72 x + y + 54z = 110.

a) 1.88

b) 1.22

c) 0

d) 1.92

View Answer

Explanation: From the given set of equations-

x=\(\frac{(85-6y+z)}{27}\)

y=\(\frac{(72-6x-2z)}{15}\)

y=\(\frac{(110-x-y)}{54}\)

1^{st} iteration:

y=0, z=0

x=\(\frac{85}{27}\)=3.14

x=3.14, z=0

y=\(\frac{72-(6×3.14)-2×(0)}{15}\)=3.54

x=3.14, y=3.54

z=\(\frac{110-3.14-3.54}{54}\)=1.91

2^{nd} iteration:

z=1.91, y=3.54

x=\(\frac{85-(6×3.54)+(1.91)}{27}\)=2.43

z=1.91, x=2.43

y=\(\frac{72-(6×2.43)-2×(1.91)}{15}\)=3.57

y=3.57, x=2.43

z=\(\frac{110-2.43-3.57}{54}\)=1.92

Thus, after the second iteration

x=2.43, y=3.57, z=1.92.

12. Solve the following equation by Gauss Seidal Method up to 3 iterations and find the value of x.

4x - 3y - z = 40 x - 6y + 2z = -28 x - 2y + 12z = -86.

a) x=11.11

b) x=13.28

c) x=11.51

d) x=9.86

View Answer

Explanation: From the given set of equations-

x=\(\frac{(40+3y+z)}{4}\)

y=\(\frac{(28+x+2z)}{6}\)

z=\(\frac{(-86-x+2y)}{12}\)

1

^{st}iteration:

y=0, z=0

x=\(\frac{40}{4}\)=10

x=10, z=0

y=\(\frac{28+(10)+2×(0)}{6}\)=6.33

x=10, y=6.33

z=\(\frac{-86-10+(2×6.33)}{12}\)=-6.94

2

^{nd}iteration:

z=-6.94, y=6.33

x=\(\frac{40+(3×6.33)+(-6.94)}{4}\)=13.01

z=-6.94, x=13.01

y=\(\frac{28+(13.01)+2×(-6.94)}{6}\)=4.52

y=4.52, x=13.01

z=\(\frac{-86-13.01+(2×4.52)}{12}\)=-7.49

3

^{rd}iteration:

z=-7.49, y=4.52

x=\(\frac{40+(3×4.52)+(-7.49)}{4}\)=11.51

z=-7.49, x=11.51

y=\(\frac{28+(11.51)+2×(-7.49)}{6}\)=4.08

y=4.08, x=11.51

z=\(\frac{-86-11.51+(2×4.08)}{12}\)=-7.44

Thus, after the third iteration

x=11.51, y=4.08, z=-7.44.

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