EDU602 Assignment 2 Solution and Discussion
MTH603 Grand Quiz Solution and Discussion

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called
Each number in the sequence is half the value of the number receding it. So the common difference in the series is dividing by two.
16÷2=8
8÷2=4
4÷2=2
2÷2=1
1÷2=½
The answer is ½ or 0.5
When you keep dividing by two, you will notice an interesting pattern: the denominator continues to increase by two, while the numerator value remains the same. That’s fascinating because in natural, whole numbers the numbers in the series would decrease by two.
1/4 , 1/8 , 1/16 etc.

A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called

More MCQs for Numerical Analysis click here

@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. 
 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

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
 Solve the system of equations by Jacobi’s iteration method.
10x = y – x = 11.19
x + 10y + z = 28.08
x + y + 10z = 35.61correct 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 = 3.45Answer: d
Explanation: Rewriting the equations as,
x = 110 (11.19 – y + z)
y = 110 (28.08 – x – z)
z = 110 (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 = 110 (11.19 – 2.808 + 3.651) = 1.19
y = 110 (28.08 – 1.119 – 3.561) = 2.34
z = 110 (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. 
 Solve the system of equations by Jacobi’s iteration method.
10x = y – x = 11.19
x + 10y + z = 28.08
x + y + 10z = 35.61correct 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 
@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
 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 = 1Answer: a
Explanation: We write the equations in the form
x = 120 (17 – y +2z)
y = 120 (18 3x + z)
z = 120 (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.  Solve the system of equations by Jacobi’s iteration method.

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

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
 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.  Which of the following is another name for Jacobi’s method?

 Which of the following is another name for Jacobi’s method?
a) Displacement method
b) Simultaneous displacement method
c) Simultaneous method
d) Diagonal method
 Which of the following is another name for Jacobi’s method?

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

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
 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.  How many assumptions are there in Jacobi’s method?

 How many assumptions are there in Jacobi’s method?
a) 2
b) 3
c) 4
d) 5
 How many assumptions are there in Jacobi’s method?

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:
 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 (nondiagonal) entries in that row. This helps in converging the result and hence it is an assumption.  Which of the following is an assumption of Jacobi’s method?

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