Skip to content
Calcrivo

Gauss-Seidel Calculator

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

Inputs

x₁

1.04326923

x₂

2.26923077

x₃

-1.08173077

Iterations

8

Max residual

8.7408 × 10⁻¹⁰

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. Gauss-Seidel iteration

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

  3. x₁

    = 1.04326923

  4. x₂

    = 2.26923077

  5. x₃

    = -1.08173077

  6. Iterations

    = 8

  7. Max residual

    = 0.0000

  8. Status

    = Converged

How it works

Gauss-Seidel improves on Jacobi by using updated values immediately: when computing xᵢ^(k+1), it uses xⱼ^(k+1) for j < i (already computed this iteration) and xⱼ^(k) for j > i. This typically converges faster than Jacobi. Guaranteed to converge for diagonally dominant or symmetric positive definite matrices.

Formula

Gauss-Seidel iteration

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

xⱼ^(k+1)
Already-updated values (this iteration)
xⱼ^(k)
Previous iteration values

Frequently Asked Questions

Why is Gauss-Seidel faster than Jacobi?

By using the most recent values immediately, Gauss-Seidel propagates corrections faster through the system, typically converging in fewer iterations.

Can I use any initial guess?

Yes, but a good initial guess (close to the solution) reduces iteration count. The zero vector is a common default.

You might also need