Quiz Chapter 04.08: Gauss-Seidel Method – Holistic Numerical Methods

MULTIPLE CHOICE TEST

GAUSS-SEIDEL METHOD

SIMULTANEOUS LINEAR EQUATIONS

Pick the most appropriate answer

1. A square matrix $\left[ A \right]_{n \times n}$ is diagonally dominant if

\left| a_{ii} \right| \geq \displaystyle\sum_{_{i \neq j}^{j=1}}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

\left| a_{ii} \right| \geq \displaystyle\sum_{_{i \neq j}^{j=1}}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

and

\left| a_{ii} \right| > \displaystyle\sum_{_{i \neq j}^{j=1}}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

\left| a_{ii} \right| \geq \displaystyle\sum_{j=1}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

and

\left| a_{ii} \right| > \displaystyle\sum_{j=1}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

\left| a_{ii} \right| \geq \displaystyle\sum_{j=1}^{n} \left| a_{ij} \right| , \, i=1,2,...,n

2. Using $\begin{bmatrix} x_{1}&x_{2}&x_{3} \\ \end{bmatrix} = \begin{bmatrix} 1&3&5 \\ \end{bmatrix}$ as the initial guess, the value of $\begin{bmatrix} x_{1}&x_{2}&x_{3} \\ \end{bmatrix}$ after three iterations of Gauss-Seidal method is

- - $\begin{bmatrix} 12&7&3 \\ 1&5&1 \\ 2&7&-11 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -5 \\ 6 \\ \end{bmatrix}$

\begin{bmatrix} -2.8333&-1.4333&-1.9727 \\ \end{bmatrix}

\begin{bmatrix} 1.4959&-0.90464&-0.84914 \\ \end{bmatrix}

\begin{bmatrix} 0.90666&-1.0115&-1.0242 \\ \end{bmatrix}

\begin{bmatrix} 1.2148&-0.72060&-0.82451\\ \end{bmatrix}

3. To ensure that the following system of equations,

- - $2x_{1} + 7x_{2} - 11x_{3} = 6$
  - $x_{1} + 2x_{2} + x_{3} = -5$
  - $7x_{1} + 5x_{2} + 2x_{3} = 17$

converges using the Gauss Seidal method, one can rewrite the above equations as follows:

\begin{bmatrix} 2&7&-11 \\ 1&2&1 \\ 7&5&2 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 6 \\ -5 \\ 17 \\ \end{bmatrix}

\begin{bmatrix} 7&5&2 \\ 1&2&1 \\ 2&7&-11 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 17 \\ -5 \\ 6 \\ \end{bmatrix}

\begin{bmatrix} 7&5&2 \\ 1&2&1 \\ 2&7&-11 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 6 \\ -5 \\ 17 \\ \end{bmatrix}

The equations cannot be rewritten in a form to ensure convergence.

4. For $\begin{bmatrix} 12&7&3 \\ 1&5&1 \\ 2&7&-11 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 22 \\ 7 \\ -2 \\ \end{bmatrix}$ and using $\begin{bmatrix} x_{1}&x_{2}&x_{3} \\ \end{bmatrix} = \begin{bmatrix} 1&2&1 \\ \end{bmatrix}$ as the initial guess, the value of $\begin{bmatrix} x_{1}&x_{2}&x_{3} \\ \end{bmatrix}$ are found at the end of each iteration as

Iteration #	$x_{1}$	$x_{2}$	$x_{3}$
$1$	$0.41666$	$1.1166$	$0.96818$
$2$	$0.93989$	$1.0183$	$1.0007$
$3$	$0.98908$	$1.0020$	$0.99930$
$4$	$0.99898$	$1.0003$	$1.0000$

At what first iteration number would you trust at least $1$ significant digit in your solution?

1

2

3

4

5. The algorithm for the Gauss-Seidal method to solve [A][X] = [C] is given as follows when using nmax iterations. The initial value of [X] is stored in [X].

Sub Seidal(n, a, x, rhs, nmax)
- For k = 1 To nmax
  - For i = 1 To n
    - For j = 1 to n
      - If (i <> j) Then
        
        Sum = Sum + a(i, j) * x(j)
      - endif
    - Next j
    - x(i) = (rhs(i) – Sum) / a(i, i)
  - Next i
- Next k
End Sub

Sub Seidal(n, a, x, rhs, nmax)
- For k = 1 To nmax
  - For i = 1 To n
    - Sum = 0
    - For j = 1 To n
      - If (i <> j) Then
        
        Sum = Sum + a(i, j) * x(j)
      - endif
    - Next j
    - x(i) = (rhs(i) – Sum) / a(i, i)
  - Next i
- Next k
End Sub

Sub Seidal(n, a, x, rhs, nmax)
- For k = 1 To nmax
  - For i = 1 To n
    - Sum = 0
    - For j = 1 To n
      - Sum = Sum + a(i, j) * x(j)
    - Next j
    - x(i) = (rhs(i) – Sum) / a(i, i)
  - Next i
- Next k
End Sub

Sub Seidal(n, a, x, rhs, nmax)
- For k = 1 To nmax
  - For i = 1 To n
    - Sum = 0
    - For j = 1 To n
      - If (i <> j) Then
        
        Sum = Sum + a(i, j) * x(j)
      - endif
    - Next j
    - x(i) = rhs(i) / a(i, i)
  - Next i
- Next k
End Sub

6. Thermistors measure temperature, have a nonlinear output and are valued for a limited range. So when a thermistor is manufactured, the manufacturer supplies a resistance vs. temperature curve. An accurate representation of the curve is generally given by

- - $\dfrac{1}{T} = a_{0} + a_{1} \ln \left( R \right) + a_{2} \left[ \ln \left( R \right) \right]^{2} + a_{3} \left[ \ln \left( R \right) \right]^{3}$

Where $T$ is temperature in Kelvin, $R$ is resistance in ohms, and $a_{0},a_{1},a_{2},a_{3}$ are constants of the calibration curve. Given the following for a thermistor

$R$ (ohms)	$T \left( ^{\circ} C \right)$
$1101.0$	$25.113$
$911.3$	$30.131$
$636.0$	$40.120$
$451.1$	$50.128$

the value of the temperature in $^{\circ}C$ for a measured resistance of $900$ ohms most nearly is

30.002

30.472

31.272

31.445