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UNIT 1 • LINEAR SYSTEMS & ITERATIONS

Gauss-Seidel Method
Iterative System Solver

The Gauss-Seidel Method is an iterative technique designed to solve large systems of linear equations. By immediately plugging newly calculated variable values into subsequent equations within the exact same pass, it accelerates convergence!

SYSTEM FORMULATIONDIAGONAL DOMINANCE

\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}

1. Diagonal DominanceEnsure

|a_1| > |b_1| + |c_1|

for guaranteed convergence.

2. Isolate VariablesSolve equation 1 for

x

, 2 for

y

, 3 for

z

3. Immediate UpdateUse new

x^{(k+1)}

immediately when solving for

y^{(k+1)}

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Solve the 3x3 linear system starting from initial guess $(x_0, y_0, z_0) = (0, 0, 0)$ :

\begin{cases} 4x + y + z = 12 \\ x + 5y + 2z = 27 \\ 2x + y + 6z = 18 \end{cases}

STEP 01 • ISOLATE DIAGONAL VARIABLES

Solve each equation for its diagonal dominant variable:

x = \frac{12 - y - z}{4}

y = \frac{27 - x - 2z}{5}

z = \frac{18 - 2x - y}{6}

STEP 02 • FIRST ITERATION (IMMEDIATE REUSE)

Substitute initial guess $y = 0, z = 0$ to find $x_1$ , then immediately use $x_1$ to find $y_1$ , and so on:

x_1 = \frac{12 - 0 - 0}{4} = 3.0000

y_1 = \frac{27 - 3.0000 - 2(0)}{5} = \frac{24}{5} = 4.8000

z_1 = \frac{18 - 2(3.0000) - 4.8000}{6} = \frac{7.2}{6} = 1.2000

STEP 03 • SECOND ITERATION

Plug $(y_1, z_1) = (4.8, 1.2)$ into the recurrence formulas:

x_2 = \frac{12 - 4.8000 - 1.2000}{4} = \frac{6.0}{4} = 1.5000

y_2 = \frac{27 - 1.5000 - 2(1.2000)}{5} = \frac{23.1}{5} = 4.6200

z_2 = \frac{18 - 2(1.5000) - 4.6200}{6} = \frac{10.38}{6} = 1.7300

STEP 04 • ITERATION SUMMARY TABLE

Step (k)	x	y	z
0	0.0000	0.0000	0.0000
1	3.0000	4.8000	1.2000
2	1.5000	4.6200	1.7300
3	1.4125	4.4255	1.7916
4	1.4457	4.3742	1.8057

FINAL CONCLUSION

Converged Solution: (x, y, z) ≈ (1.45, 4.36, 1.80)

By immediately reusing newly solved values within each pass, Gauss-Seidel converges much faster than Jacobi iteration!

LABORATORY

Interactive Live Calculator & Visualizer

LINEAR ALGEBRA SYSTEM SOLVER

Gauss-Seidel Interactive Engine

PreviousNewton-Raphson Method

Unit 1•5 of 28

NextNewton Forward Interpolation

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

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UNIT 1 • LINEAR SYSTEMS & ITERATIONS

Gauss-Seidel Method
Iterative System Solver

SYSTEM FORMULATIONDIAGONAL DOMINANCE

\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}

1. Diagonal DominanceEnsure

|a_1| > |b_1| + |c_1|

for guaranteed convergence.

2. Isolate VariablesSolve equation 1 for

x

, 2 for

y

, 3 for

z

3. Immediate UpdateUse new

x^{(k+1)}

immediately when solving for

y^{(k+1)}

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Solve the 3x3 linear system starting from initial guess $(x_0, y_0, z_0) = (0, 0, 0)$ :

\begin{cases} 4x + y + z = 12 \\ x + 5y + 2z = 27 \\ 2x + y + 6z = 18 \end{cases}

STEP 01 • ISOLATE DIAGONAL VARIABLES

Solve each equation for its diagonal dominant variable:

x = \frac{12 - y - z}{4}

y = \frac{27 - x - 2z}{5}

z = \frac{18 - 2x - y}{6}

STEP 02 • FIRST ITERATION (IMMEDIATE REUSE)

Substitute initial guess $y = 0, z = 0$ to find $x_1$ , then immediately use $x_1$ to find $y_1$ , and so on:

x_1 = \frac{12 - 0 - 0}{4} = 3.0000

y_1 = \frac{27 - 3.0000 - 2(0)}{5} = \frac{24}{5} = 4.8000

z_1 = \frac{18 - 2(3.0000) - 4.8000}{6} = \frac{7.2}{6} = 1.2000

STEP 03 • SECOND ITERATION

Plug $(y_1, z_1) = (4.8, 1.2)$ into the recurrence formulas:

x_2 = \frac{12 - 4.8000 - 1.2000}{4} = \frac{6.0}{4} = 1.5000

y_2 = \frac{27 - 1.5000 - 2(1.2000)}{5} = \frac{23.1}{5} = 4.6200

z_2 = \frac{18 - 2(1.5000) - 4.6200}{6} = \frac{10.38}{6} = 1.7300

STEP 04 • ITERATION SUMMARY TABLE

Step (k)	x	y	z
0	0.0000	0.0000	0.0000
1	3.0000	4.8000	1.2000
2	1.5000	4.6200	1.7300
3	1.4125	4.4255	1.7916
4	1.4457	4.3742	1.8057

FINAL CONCLUSION

Converged Solution: (x, y, z) ≈ (1.45, 4.36, 1.80)

By immediately reusing newly solved values within each pass, Gauss-Seidel converges much faster than Jacobi iteration!

LABORATORY

Interactive Live Calculator & Visualizer

LINEAR ALGEBRA SYSTEM SOLVER

Gauss-Seidel Interactive Engine

PreviousNewton-Raphson Method

Unit 1•5 of 28

NextNewton Forward Interpolation

Gauss-Seidel Method Iterative System Solver

Step-by-Step Worked Example

Converged Solution: (x, y, z) ≈ (1.45, 4.36, 1.80)

Interactive Live Calculator & Visualizer

Gauss-Seidel Interactive Engine

NETZ Engine

Gauss-Seidel Method Iterative System Solver

Step-by-Step Worked Example

Converged Solution: (x, y, z) ≈ (1.45, 4.36, 1.80)

Interactive Live Calculator & Visualizer

Gauss-Seidel Interactive Engine

Gauss-Seidel Method
Iterative System Solver

Gauss-Seidel Method
Iterative System Solver