289 lines
9.0 KiB
Typst
289 lines
9.0 KiB
Typst
#import "template/module.typ": *
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#show: default.with(
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title: "Numerical Methods Assignment 2",
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)
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= Gaussian Elimination with Partial Pivoting (15%)
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Solve the following system by Gaussian elimination with partial pivoting:
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$ mat(4, 2, -2, -1; 0, 4, 1, 2; 3, -2, 1, 2; 2, 0, 3, -5) vec(x_1, x_2, x_3, x_4) = vec(7, 10, 2, 3) $
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== Solution Process
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*Initial augmented matrix:*
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$ mat(augment: #4, 4, 2, -2, -1, 7; 0, 4, 1, 2, 10; 3, -2, 1, 2, 2; 2, 0, 3, -5, 3) $
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*Step 1 - Column 1:* The pivot element is $a_(11) = 4$. Checking column 1: $|4| >= |0|, |3|, |2|$. No row interchange needed.
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Multipliers: $m_(21) = 0$, $m_(31) = 3/4 = 0.75$, $m_(41) = 2/4 = 0.5$
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After elimination:
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$ mat(augment: #4, 4, 2, -2, -1, 7; 0, 4, 1, 2, 10; 0, -3.5, 2.5, 2.75, -3.25; 0, -1, 4, -4.5, -0.5) $
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*Step 2 - Column 2:* The pivot element is $a_(22) = 4$. Checking: $|4| >= |-3.5|, |-1|$. No row interchange needed.
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Multipliers: $m_(32) = -3.5/4 = -0.875$, $m_(42) = -1/4 = -0.25$
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After elimination:
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$ mat(augment: #4, 4, 2, -2, -1, 7; 0, 4, 1, 2, 10; 0, 0, 3.375, 4.5, 5.5; 0, 0, 4.25, -4, 2) $
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*Step 3 - Column 3:* Checking: $|3.375| < |4.25|$. *Row interchange required: swap row 3 with row 4.*
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After interchange:
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$ mat(augment: #4, 4, 2, -2, -1, 7; 0, 4, 1, 2, 10; 0, 0, 4.25, -4, 2; 0, 0, 3.375, 4.5, 5.5) $
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Multiplier: $m_(43) = 3.375/4.25 = 0.7941$
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After elimination:
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$ mat(augment: #4, 4, 2, -2, -1, 7; 0, 4, 1, 2, 10; 0, 0, 4.25, -4, 2; 0, 0, 0, 7.6765, 3.9118) $
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*Back substitution:*
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- $x_4 = 3.9118 / 7.6765 = 0.509579$
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- $x_3 = (2 - (-4)(0.509579)) / 4.25 = 0.950192$
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- $x_2 = (10 - (1)(0.950192) - (2)(0.509579)) / 4 = 2.007663$
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- $x_1 = (7 - (2)(2.007663) - (-2)(0.950192) - (-1)(0.509579)) / 4 = 1.348659$
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== Solution
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$ x_1 = 1.348659, quad x_2 = 2.007663, quad x_3 = 0.950192, quad x_4 = 0.509579 $
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*Column requiring row interchange:* Column 3
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#pagebreak()
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= Scaled Partial Pivoting (25%)
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Solve the system using scaled partial pivoting:
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$ mat(4.13, -2.20, 0.95, 3.02; 6.14, 4.45, -1.45, -4.02; 1.03, 1.86, 0.44, 5.22) $
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== Part (a): Six Significant Digits
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*Scale factors:*
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- $s_1 = max(|4.13|, |-2.20|, |0.95|) = 4.13$
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- $s_2 = max(|6.14|, |4.45|, |-1.45|) = 6.14$
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- $s_3 = max(|1.03|, |1.86|, |0.44|) = 1.86$
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*Step 1 - Column 1:*
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Scaled ratios:
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- Row 1: $|4.13|/4.13 = 1.000$
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- Row 2: $|6.14|/6.14 = 1.000$
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- Row 3: $|1.03|/1.86 = 0.554$
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Rows 1 and 2 are tied; keep current order.
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Multipliers: $m_(21) = 6.14/4.13 = 1.48668$, $m_(31) = 1.03/4.13 = 0.249395$
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After elimination:
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$ mat(augment: #3, 4.13, -2.20, 0.95, 3.02; 0, 7.7207, -2.86235, -8.50977; 0, 2.40867, 0.203075, 4.46683) $
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*Step 2 - Column 2:*
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Scaled ratios:
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- Row 2: $|7.7207|/6.14 = 1.257$
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- Row 3: $|2.40867|/1.86 = 1.295$
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Row 3 has larger ratio. *Interchange rows 2 and 3.*
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Multiplier: $m_(22) = 7.7207/2.40867 = 3.20538$
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After elimination:
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$ mat(augment: #3, 4.13, -2.20, 0.95, 3.02; 0, 2.40867, 0.203075, 4.46683; 0, 0, -3.51328, -22.8277) $
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*Back substitution:*
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- $x_3 = -22.8277 / (-3.51328) = 6.49755$
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- $x_2 = (4.46683 - 0.203075 times 6.49755) / 2.40867 = 1.30667$
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- $x_1 = (3.02 - 0.95 times 6.49755 - (-2.20) times 1.30667) / 4.13 = -0.0673$
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*Solution (6 significant digits):*
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$ x_1 = -0.0673, quad x_2 = 1.3067, quad x_3 = 6.4976 $
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== Part (b): Three Significant Digits
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Following the same process with 3-digit arithmetic:
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*Solution (3 significant digits):*
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$ x_1 = -0.0688, quad x_2 = 1.31, quad x_3 = 6.51 $
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== Comparison
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The solutions are close but show small differences due to rounding errors:
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- $x_1$: $-0.0673$ vs $-0.0688$ (difference $approx 2%$)
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- $x_2$: $1.3067$ vs $1.31$ (difference $approx 0.3%$)
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- $x_3$: $6.4976$ vs $6.51$ (difference $approx 0.2%$)
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The solution with 3 significant digits is slightly different but still reasonable. The accumulated rounding errors affect $x_1$ the most since it is computed last in back substitution.
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#pagebreak()
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= Symmetric Tridiagonal System (30%)
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The system:
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$ mat(4, -1, 0, 0, 0, 0; -1, 4, -1, 0, 0, 0; 0, -1, 4, -1, 0, 0; 0, 0, -1, 4, -1, 0; 0, 0, 0, -1, 4, -1; 0, 0, 0, 0, -1, 4) vec(x_1, x_2, x_3, x_4, x_5, x_6) = vec(100, 200, 200, 200, 200, 100) $
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== Part (a): Algorithm for Symmetric Tridiagonal System
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For a symmetric tridiagonal matrix, we only need to store:
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- $d_i$: diagonal elements ($n$ elements)
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- $a_i$: off-diagonal elements ($n-1$ elements, same above and below diagonal)
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- $b_i$: right-hand side ($n$ elements)
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*Algorithm (Modified Thomas Algorithm for Symmetric Case):*
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```
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// Input: d[1..n] (diagonal), a[1..n-1] (off-diagonal), b[1..n] (RHS)
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// Output: x[1..n] (solution)
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// Forward Elimination
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for i = 2 to n:
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m = a[i-1] / d[i-1]
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d[i] = d[i] - m * a[i-1]
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b[i] = b[i] - m * b[i-1]
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// Back Substitution
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x[n] = b[n] / d[n]
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for i = n-1 downto 1:
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x[i] = (b[i] - a[i] * x[i+1]) / d[i]
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```
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== Part (b): Solving the System
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*Initial compact representation:*
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#align(center)[
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#table(
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columns: 4,
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[Row], [Sub-diagonal], [Diagonal], [RHS],
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[1], [-], [4], [100],
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[2], [-1], [4], [200],
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[3], [-1], [4], [200],
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[4], [-1], [4], [200],
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[5], [-1], [4], [200],
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[6], [-1], [4], [100],
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)
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]
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*Forward Elimination:*
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- Step 1: $m = -1/4 = -0.25$, $d'_2 = 4 - (-0.25)(-1) = 3.75$, $b'_2 = 200 - (-0.25)(100) = 225$
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- Step 2: $m = -1/3.75 = -0.2667$, $d'_3 = 3.7333$, $b'_3 = 260$
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- Step 3: $m = -1/3.7333 = -0.2679$, $d'_4 = 3.7321$, $b'_4 = 269.64$
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- Step 4: $m = -1/3.7321 = -0.2679$, $d'_5 = 3.7321$, $b'_5 = 272.25$
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- Step 5: $m = -1/3.7321 = -0.2679$, $d'_6 = 3.7321$, $b'_6 = 172.95$
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*Back Substitution:*
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- $x_6 = 172.95/3.7321 = 46.34$
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- $x_5 = (272.25 - (-1)(46.34))/3.7321 = 85.37$
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- $x_4 = (269.64 - (-1)(85.37))/3.7321 = 95.12$
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- $x_3 = (260 - (-1)(95.12))/3.7333 = 95.12$
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- $x_2 = (225 - (-1)(95.12))/3.75 = 85.37$
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- $x_1 = (100 - (-1)(85.37))/4 = 46.34$
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*Solution:*
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$ x_1 = 46.34, quad x_2 = 85.37, quad x_3 = 95.12, quad x_4 = 95.12, quad x_5 = 85.37, quad x_6 = 46.34 $
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== Part (c): Arithmetic Operations Count
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For a system of $N$ equations:
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*Forward Elimination (loop runs $N-1$ times):*
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- $N-1$ divisions (computing $m$)
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- $N-1$ multiplications (computing $m dot a_(i-1)$)
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- $2(N-1)$ subtractions (updating $d_i$ and $b_i$)
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Total: $4(N-1)$ operations
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*Back Substitution:*
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- 1 division (computing $x_N$)
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- For $i = N-1$ down to 1: 1 multiplication + 1 subtraction + 1 division = $3(N-1)$ operations
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Total: $1 + 3(N-1) = 3N - 2$ operations
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*Total operations:* $4(N-1) + 3N - 2 = 7N - 6 = O(N)$
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This is much more efficient than general Gaussian elimination which requires $O(N^3)$ operations.
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#pagebreak()
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= Jacobi and Gauss-Seidel Methods (40%)
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Original system:
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$ mat(7, -3, 4; -3, 2, 6; 2, 5, 3) vec(x_1, x_2, x_3) = vec(6, 2, -5) $
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== Checking and Rearranging for Diagonal Dominance
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*Original matrix diagonal dominance check:*
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- Row 1: $|7| = 7$, $|-3| + |4| = 7$ (equal, not strictly dominant)
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- Row 2: $|2| = 2$, $|-3| + |6| = 9$ (NOT dominant)
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- Row 3: $|3| = 3$, $|2| + |5| = 7$ (NOT dominant)
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*Rearrangement:* Swap equations 2 and 3:
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$ mat(7, -3, 4; 2, 5, 3; -3, 2, 6) vec(x_1, x_2, x_3) = vec(6, -5, 2) $
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After rearrangement:
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- Row 1: $|7| = 7$, $|-3| + |4| = 7$ (weakly dominant)
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- Row 2: $|5| = 5$, $|2| + |3| = 5$ (weakly dominant)
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- Row 3: $|6| = 6 > |-3| + |2| = 5$ (strictly dominant)
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The system is weakly diagonally dominant. The methods may still converge.
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== Part (a): Jacobi Method
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*Iteration formulas:*
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$ x_1^((k+1)) &= (6 + 3x_2^((k)) - 4x_3^((k))) / 7 \
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x_2^((k+1)) &= (-5 - 2x_1^((k)) - 3x_3^((k))) / 5 \
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x_3^((k+1)) &= (2 + 3x_1^((k)) - 2x_2^((k))) / 6 $
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*Starting vector:* $bold(x)^((0)) = [0, 0, 0]^T$
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#align(center)[
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#table(
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columns: 4,
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[Iteration], [$x_1$], [$x_2$], [$x_3$],
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[0], [0], [0], [0],
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[1], [0.8571], [-1.0000], [0.3333],
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[2], [0.2381], [-1.5429], [1.0952],
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[3], [-0.4299], [-1.7524], [0.9667],
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[4], [-0.4463], [-1.4080], [0.7025],
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[5], [-0.1477], [-1.2430], [0.5795],
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[...], [...], [...], [...],
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[37], [-0.1433], [-1.3746], [0.7199],
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)
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]
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*Jacobi converged after 37 iterations.*
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*Solution:* $x_1 = -0.1433, quad x_2 = -1.3746, quad x_3 = 0.7199$
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== Part (b): Gauss-Seidel Method
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*Iteration formulas (using most recent values):*
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$ x_1^((k+1)) &= (6 + 3x_2^((k)) - 4x_3^((k))) / 7 \
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x_2^((k+1)) &= (-5 - 2x_1^((k+1)) - 3x_3^((k))) / 5 \
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x_3^((k+1)) &= (2 + 3x_1^((k+1)) - 2x_2^((k+1))) / 6 $
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#align(center)[
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#table(
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columns: 4,
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[Iteration], [$x_1$], [$x_2$], [$x_3$],
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[0], [0], [0], [0],
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[1], [0.8571], [-1.3429], [1.2095],
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[2], [-0.4095], [-1.5619], [0.6492],
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[3], [-0.1832], [-1.3162], [0.6805],
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[4], [-0.0958], [-1.3700], [0.7421],
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[5], [-0.1540], [-1.3836], [0.7175],
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[...], [...], [...], [...],
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[16], [-0.1433], [-1.3746], [0.7199],
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)
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]
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*Gauss-Seidel converged after 16 iterations.*
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*Solution:* $x_1 = -0.1433, quad x_2 = -1.3746, quad x_3 = 0.7199$
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== Comparison
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Both methods converge to the same solution:
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$ x_1 approx -0.143, quad x_2 approx -1.375, quad x_3 approx 0.720 $
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Gauss-Seidel converged in *16 iterations*, while Jacobi required *37 iterations*. This is expected because Gauss-Seidel uses updated values immediately, leading to faster convergence.
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