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Calcrivo

Jacobi Method Calculator

Solve a diagonally-dominant linear system iteratively using the Jacobi method.

Inputs

x₁

1.04326923

x₂

2.26923077

x₃

-1.08173077

Iterations

15

Max residual

8.7240 × 10⁻⁹

Status

Converged

Step by step

  1. Values used

    a₁₁ = 10; a₁₂ = -1; a₁₃ = 2; b₁ = 6; a₂₁ = -1; a₂₂ = 11; a₂₃ = -1; b₂ = 25; a₃₁ = 2; a₃₂ = -1; a₃₃ = 10; b₃ = -11; Tolerance = 0.0000; Max iterations = 100

  2. Jacobi iteration

    xᵢ^(k+1) = (bᵢ - Σⱼ≠ᵢ aᵢⱼ·xⱼ^(k)) / aᵢᵢ

  3. x₁

    = 1.04326923

  4. x₂

    = 2.26923077

  5. x₃

    = -1.08173077

  6. Iterations

    = 15

  7. Max residual

    = 0.0000

  8. Status

    = Converged

How it works

The Jacobi method solves Ax = b iteratively by isolating each variable: xᵢ^(k+1) = (bᵢ − Σⱼ≠ᵢ aᵢⱼxⱼ^(k)) / aᵢᵢ. All updates use values from the previous iteration (unlike Gauss-Seidel). Convergence is guaranteed for diagonally dominant systems. The method reports iteration count and achieved residual.

Formula

Jacobi iteration

xᵢ^(k+1) = (bᵢ - Σⱼ≠ᵢ aᵢⱼ·xⱼ^(k)) / aᵢᵢ

xᵢ^(k+1)
New value of variable i
xⱼ^(k)
Previous iteration values
aᵢⱼ
Matrix coefficients

Frequently Asked Questions

When does Jacobi converge?

Convergence is guaranteed when the matrix is strictly diagonally dominant: |aᵢᵢ| > Σⱼ≠ᵢ |aᵢⱼ| for all rows.

How does Jacobi compare to Gauss-Seidel?

Gauss-Seidel uses updated values immediately (faster convergence) but is sequential. Jacobi can be parallelized since it uses only previous-iteration values.

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