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    MTH603 Grand Quiz Solution and Discussion

    MTH603 - Numerical Analysis
    mth603 grand quiz solution discussion spring 2020
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    • zaasmi
      zaasmi Cyberian's Gold last edited by

      Grand Quiz Total Questions : 30
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      • zaasmi
        zaasmi Cyberian's Gold @zaasmi last edited by

        @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.

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        • zaasmi
          zaasmi Cyberian's Gold last edited by

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

          Discussion is right way to get Solution of the every assignment, Quiz and GDB.
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          • zaasmi
            zaasmi Cyberian's Gold last edited by

            More MCQs for Numerical Analysis click here

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            • zaasmi
              zaasmi Cyberian's Gold @zaasmi last edited by

              @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

              1. 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.

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              • zaasmi
                zaasmi Cyberian's Gold last edited by

                1. 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

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                zaasmi 1 Reply Last reply Reply Quote 0
                • zaasmi
                  zaasmi Cyberian's Gold @zaasmi last edited by

                  @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

                  1. 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 = 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.

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                  • zaasmi
                    zaasmi Cyberian's Gold last edited by

                    1. 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

                    Discussion is right way to get Solution of the every assignment, Quiz and GDB.
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                    zaasmi 1 Reply Last reply Reply Quote 0
                    • zaasmi
                      zaasmi Cyberian's Gold @zaasmi last edited by

                      @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

                      1. 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 = 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.

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                      • zaasmi
                        zaasmi Cyberian's Gold last edited by zaasmi

                        1. 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

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                        • zaasmi
                          zaasmi Cyberian's Gold @zaasmi last edited by

                          @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

                          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.

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                          • zaasmi
                            zaasmi Cyberian's Gold last edited by

                            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

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                            • zaasmi
                              zaasmi Cyberian's Gold last edited by

                              1. 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.

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                              • zaasmi
                                zaasmi Cyberian's Gold @zaasmi last edited by

                                @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

                                1. 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.

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                                • zaasmi
                                  zaasmi Cyberian's Gold last edited by

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

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                                  • zaasmi
                                    zaasmi Cyberian's Gold @zaasmi last edited by

                                    @zaasmi said in MTH603 Grand Quiz Solution and Discussion:

                                    1. 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.

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