How this instrument works
Gauss-Jordan elimination reduces a system of equations all the way until each variable stands completely alone on its own row, with a coefficient of exactly 1 and nothing else on that row's left-hand side — at which point the solution can simply be read straight off the constants, no separate back-substitution step required.
This site's companion Row Echelon Form calculator demonstrates the FIRST elimination step toward that goal, clearing row 2's leading variable using row 1; Gauss-Jordan elimination is that same idea carried all the way through, both rows fully reduced rather than just one partially cleared one.
For a 2-by-2 system, the fully reduced solution has a direct closed form via Cramer's Rule — x and y each come out as a ratio of two determinants, sidestepping the row-by-row elimination process entirely while landing on the identical answer.
- Enter row 1's x coefficient, y coefficient, and constant.
- Enter row 2's x coefficient, y coefficient, and constant.
- Read Solution: x and Solution: y — both variables, fully isolated.
- Substitute both values back into the original two equations to confirm they satisfy both simultaneously.
Worked example — solving [2x+y=5; x+3y=5]
For 2x+y=5 and x+3y=5: fully reducing both rows gives x=2 and y=1 directly, with no back-substitution needed. Checking both original equations: 2×2+1=5 ✓ and 2+3×1=5 ✓, confirming the solution satisfies both simultaneously.
For [x+0y=4; 0x+y=7], the system arrives already fully reduced — reading the solution straight off the constants gives x=4, y=7 immediately. For [x+y=10; x−y=2]: fully reducing gives x=6, y=4, checked by 6+4=10 ✓ and 6−4=2 ✓.
Questions
What is Gauss-Jordan elimination?
A method for solving a system of equations by reducing every row until each variable is fully isolated with a coefficient of exactly 1, letting the solution be read directly off the resulting constants without any separate back-substitution.
How is this different from ordinary Gaussian elimination (row echelon form)?
Ordinary row reduction stops once the system reaches row echelon form, with some rows still needing back-substitution to fully solve; Gauss-Jordan elimination continues further, clearing every row completely so no back-substitution step remains.
What if the system has no unique solution?
That happens whenever the two equations are parallel or identical — the determinant ad−bc comes out to exactly 0, and no single (x,y) pair satisfies both equations uniquely.
How does this relate to Cramer's Rule?
The closed-form solution used here, x and y each as a ratio of two determinants, IS Cramer's Rule — a direct shortcut to the same answer full Gauss-Jordan row reduction would eventually reach, just without carrying out the row operations explicitly.
How can I check my solution is correct?
Substitute the found x and y values back into BOTH original equations — a genuine solution must satisfy both simultaneously, not just one of them.