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]]>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.

]]>

- 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.

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

- 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

Answer: 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.

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

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

20x + y – 2z = 17

3x + 20y – z = -18

2x – 3y + 20z = 25a) 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 = 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.

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

- 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.

a) Displacement method

b) Simultaneous displacement method

c) Simultaneous method

d) Diagonal method

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.

- 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.

a) 2

b) 3

c) 4

d) 5

- 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.

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

- 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.

a) True

b) False

- 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

View Answer

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.

a) Leading diagonal

b) Last column

c) Last row

d) Non-leading diagonal

View Answer

In J

S

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

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

https://cyberian.pk/topic/838/mth603-mid-term-past-and-current-solved-paper-discussion

SLOW

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

https://cyberian.pk/topic/838/mth603-mid-term-past-and-current-solved-paper-discussion

]]>For the system of equations; x =2, y=3. The inverse of the matrix associated with its coefficients is-----------.

⌈ 3 -2⌉ ⌈ x ⌉ = ⌈ 5 ⌉

⌊ 4 3⌋ ⌊ y ⌋ ⌊ -2 ⌋

⌈ x ⌉ = (1/17) ⌈ 3 2 ⌉ ⌈ 5 ⌉

⌊ y ⌋ ⌊ -4 3 ⌋ ⌊-2⌋⌋

```
= (1/17)⌈11⌉
⌊-26⌋
= ⌈11/17⌉
⌊-26/17⌋
```

x = 11/17 and y = -26/17

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