Calxsolver
Linear Algebra

Solve a Linear System (Ax = b)

Enter the coefficient matrix A and the right-hand side vector b. The calculator builds the augmented matrix [A | b], reduces it with Gauss-Jordan elimination, and reports a unique solution, no solution, or infinitely many solutions with a parametric form.

Solve a Linear System (Ax = b)

Gauss-Jordan on the augmented matrix — unique, infinite or no solution.

Try:
Answerx1 = 1, x2 = 3
  1. SystemA is 2×2, b has 2 entries.
  2. Matrix A[[2, 1], [1, 3]]
  3. Vector b(5, 10)
  4. Augmented [A | b][[2, 1, 5], [1, 3, 10]]
  5. Row operationsR1 → (1/2)·R1; R2 → R2 − (1)·R1; R2 → (1/2.5)·R2; R1 → R1 − (0.5)·R2
  6. RREF of [A | b][[1, 0, 1], [0, 1, 3]]
  7. ConclusionUnique solution — pivot in every variable column.
  8. Solutionx1 = 1, x2 = 3
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Worked examples

Solve 2x + y = 5, x + 3y = 10 Solve a 2×2 linear system with Gauss-Jordan elimination.

Learn why this works

Vector Spaces, Span, and Basis Why linear independence determines whether a system has solutions →

Frequently asked questions

How does the calculator detect no solution?

If the reduced row echelon form contains a row of the form [0 0 … 0 | c] with c ≠ 0, the system is inconsistent and has no solution.

When are there infinitely many solutions?

When the system is consistent but has fewer pivot columns than variables — the variables without pivots are free, giving a parametric family.

Does this require a square matrix?

No. The method works for any m×n system. Over- and under-determined systems are handled equally well.

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