Solving linear systems: Gaussian elimination, determinants and inverses
Linear systems of equations — expressions like a₁₁x + a₁₂y = b₁ — appear throughout mathematics, physics, engineering and data science. Solving systems efficiently and understanding their structure (unique solution, no solution, infinite solutions) is crucial. This article explains practical methods for 2×2 and 3×3 systems: Gaussian elimination, use of determinants, and matrix inverse methods, with worked examples and numerical considerations.
Matrix representation
A linear system can be written in matrix form as A·x = b, where A is the coefficient matrix, x is the vector of unknowns, and b is the RHS vector. For a 2×2 system:
[a11 a12] [x] = [b1] [a21 a22] [y] [b2]
Matrix notation is convenient for both theory and computation. Many solution methods operate directly on A and b together.
Gaussian elimination (row reduction)
Gaussian elimination uses elementary row operations to transform the augmented matrix [A | b] into an upper-triangular (row-echelon) form, then back-substitution recovers unknowns. Elementary row operations are: swap rows, multiply a row by a non-zero scalar, and add a multiple of one row to another. These operations preserve the solution set.
Detecting special cases
During elimination you may encounter a row of zeros on the coefficient side. If the corresponding RHS is also zero, that row indicates a dependent equation (possible infinitely many solutions). If RHS ≠ 0, the system is inconsistent (no solution). A non-zero determinant of A guarantees a unique solution (for square systems).
Determinant and inverse
For 2×2 matrices, the determinant is simple: det(A) = a11·a22 − a12·a21. For 3×3 matrices, compute via Sarrus' rule or cofactor expansion. If det(A) ≠ 0, A is invertible and x = A⁻¹ b gives the unique solution. Computing the inverse explicitly is more expensive than Gaussian elimination for large systems, but for 2×2 and 3×3 it is practical and instructive.
Worked example (2×2)
Solve 2x + 3y = 7 and 4x − y = 1. Build augmented matrix and perform elimination to find x and y. The determinant is (2)(−1) − (3)(4) = −2 − 12 = −14 ≠ 0, so a unique solution exists. Solving yields x = 1, y = 5/3 (worked steps shown in calculator).
Worked example (3×3)
3×3 systems follow the same pattern: reduce to upper triangular, back-substitute. Because the number of operations increases, careful pivoting (swapping rows to use a larger pivot) improves numerical stability — though for small systems this is often optional. The online solver demonstrates row swaps, scaling, and elimination steps explicitly.
Numerical considerations
Computations in web calculators use JavaScript's floating point arithmetic, which is fast and sufficiently precise for many tasks but can introduce round-off errors when coefficients differ widely in magnitude. Partial pivoting (choosing the largest pivot in a column) reduces rounding error. For exact arithmetic with rationals use symbolic math libraries.
When to use determinant vs elimination
Determinants are useful diagnostics: non-zero determinant means unique solution. However, solving the system by computing an inverse via determinants (Cramer's rule) is inefficient for large matrices. Gaussian elimination is the standard numeric approach and scales better.
Educational vs practical use
This solver is a great educational tool — it shows each elementary row operation and intermediate augmented matrix — which helps learners understand why and how elimination works. For large-scale numerical computation, dedicated numeric libraries (LAPACK, Eigen) with robust pivoting and stability controls are preferred.
Conclusion
Understanding Gaussian elimination, determinants, and inverses forms the backbone of linear algebra. This solver provides both rapid numeric answers and detailed, step-by-step explanations to aid learning and verification.