UNIT 4 • LINEAR ALGEBRA & MATRIX METHODS

Jacobi Iterative Method
Step-by-Step Friendly Guide

The Jacobi Method is an iterative solver for systems $Ax = b$ . Starting from an initial guess (typically $x^{(0)} = 0$ ), each variable is updated simultaneously using the previous iteration's values — converging when the matrix $A$ is strictly diagonally dominant.

JACOBI UPDATE FORMULASIMULTANEOUS UPDATE

x_i^{(k+1)} = \frac{1}{a_{ii}}\left(b_i - \sum_{j \neq i} a_{ij} x_j^{(k)}\right)

All $x_i$ are updated simultaneously using values from iteration $k$ (unlike Gauss-Seidel which uses newest available values).

Convergence ConditionMatrix must be strictly diagonally dominant:

|a_{ii}| > \sum_{j \neq i}|a_{ij}|

for all rows.

Jacobi vs Gauss-SeidelJacobi uses all old values for each update; Gauss-Seidel uses fresh values immediately (faster convergence).

Parallelism AdvantageAll variable updates are independent per iteration — making Jacobi naturally parallelizable on GPUs.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM

Solve: $10x_1 - x_2 + 2x_3 = 6,\quad -x_1 + 11x_2 - x_3 = 25,\quad 2x_1 - x_2 + 10x_3 = -11$

Expected solution: x₁ = 1, x₂ = 2, x₃ = -1

STEP 01 • REARRANGE FOR EACH VARIABLE

x_1^{(k+1)} = \frac{6 + x_2^{(k)} - 2x_3^{(k)}}{10}

x_2^{(k+1)} = \frac{25 + x_1^{(k)} + x_3^{(k)}}{11}

x_3^{(k+1)} = \frac{-11 - 2x_1^{(k)} + x_2^{(k)}}{10}

STEP 02 • ITERATION 1 (starting from x = [0,0,0])

x_1^{(1)} = \frac{6 + 0 - 0}{10} = 0.6

x_2^{(1)} = \frac{25 + 0 + 0}{11} = 2.2727

x_3^{(1)} = \frac{-11 - 0 + 0}{10} = -1.1

LABORATORY

Interactive Live Calculator

ITERATIVE MATRIX LABORATORY

Jacobi Method Engine

Matrix Size:

💡 Yellow column = constants vector b. Coefficient matrix A is on the left. Full system: Ax = b.

Tolerance

Max Iterations

PreviousLU Decomposition

Unit 4•23 of 34

NextTaylor's Series Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 4 • LINEAR ALGEBRA & MATRIX METHODS

Jacobi Iterative Method
Step-by-Step Friendly Guide

JACOBI UPDATE FORMULASIMULTANEOUS UPDATE

x_i^{(k+1)} = \frac{1}{a_{ii}}\left(b_i - \sum_{j \neq i} a_{ij} x_j^{(k)}\right)

All $x_i$ are updated simultaneously using values from iteration $k$ (unlike Gauss-Seidel which uses newest available values).

Convergence ConditionMatrix must be strictly diagonally dominant:

|a_{ii}| > \sum_{j \neq i}|a_{ij}|

for all rows.

Jacobi vs Gauss-SeidelJacobi uses all old values for each update; Gauss-Seidel uses fresh values immediately (faster convergence).

Parallelism AdvantageAll variable updates are independent per iteration — making Jacobi naturally parallelizable on GPUs.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM

Solve: $10x_1 - x_2 + 2x_3 = 6,\quad -x_1 + 11x_2 - x_3 = 25,\quad 2x_1 - x_2 + 10x_3 = -11$

Expected solution: x₁ = 1, x₂ = 2, x₃ = -1

STEP 01 • REARRANGE FOR EACH VARIABLE

x_1^{(k+1)} = \frac{6 + x_2^{(k)} - 2x_3^{(k)}}{10}

x_2^{(k+1)} = \frac{25 + x_1^{(k)} + x_3^{(k)}}{11}

x_3^{(k+1)} = \frac{-11 - 2x_1^{(k)} + x_2^{(k)}}{10}

STEP 02 • ITERATION 1 (starting from x = [0,0,0])

x_1^{(1)} = \frac{6 + 0 - 0}{10} = 0.6

x_2^{(1)} = \frac{25 + 0 + 0}{11} = 2.2727

x_3^{(1)} = \frac{-11 - 0 + 0}{10} = -1.1

LABORATORY

Interactive Live Calculator

ITERATIVE MATRIX LABORATORY

Jacobi Method Engine

Matrix Size:

💡 Yellow column = constants vector b. Coefficient matrix A is on the left. Full system: Ax = b.

Tolerance

Max Iterations

PreviousLU Decomposition

Unit 4•23 of 34

NextTaylor's Series Method

Jacobi Iterative Method Step-by-Step Friendly Guide

Step-by-Step Worked Example

Interactive Live Calculator

Jacobi Method Engine

NETZ Engine

Jacobi Iterative Method Step-by-Step Friendly Guide

Step-by-Step Worked Example

Interactive Live Calculator

Jacobi Method Engine

Jacobi Iterative Method
Step-by-Step Friendly Guide

Jacobi Iterative Method
Step-by-Step Friendly Guide