Any linear system can be packed into a single matrix equation and solved in one move. This is the matrix view of solving simultaneous equations, and once you see it, three-variable systems stop being intimidating.
On this page
Writing the system as AX = B
Take the system:
x + 2y = 5
3x + 4y = 6Separate coefficients, unknowns and constants into three matrices:
A = 1 2 X = x B = 5
3 4 y 6Then the whole system is simply A X = B.
The inverse-matrix method
If A has an inverse, multiply both sides on the left by A⁻¹:
A⁻¹ A X = A⁻¹ B → X = A⁻¹ BSo the unknowns are just A⁻¹ times B. That's the entire method.
Worked 2-variable example
From the matrix pillar we know A = [[1, 2], [3, 4]] has A⁻¹ = [[−2, 1], [1.5, −0.5]]. Multiply by B = [5, 6]:
- x = (−2)(5) + (1)(6) = −10 + 6 = −4
- y = (1.5)(5) + (−0.5)(6) = 7.5 − 3 = 4.5
x + 2y = −4 + 9 = 5 ✔ and 3x + 4y = −12 + 18 = 6 ✔
To do this on the calculator, enter A in MATRIX mode to get A⁻¹, then apply it to B.
The faster route: EQN mode
For most homework, you don't even need the inverse. Switch to EQN mode, choose simultaneous equations, and enter the coefficients and constants directly — the calculator returns x and y (and z for three unknowns) in one step. See systems of equations with 2–3 unknowns for that workflow, and the equation solver pillar for the full picture.
When it fails
If the coefficient matrix is singular, there's no unique solution — the equations are either contradictory (no solution) or dependent (infinitely many). The inverse method only works when det(A) ≠ 0.
Frequently asked questions
How do matrices solve simultaneous equations?
Write AX = B; if A is invertible, X = A⁻¹B gives the unknowns directly.
What if A isn't invertible?
det(A) = 0 means no unique solution — either none or infinitely many.
Can it solve 3-variable systems?
Yes — EQN mode handles 2–3 unknowns directly, or use a 3×3 inverse in MATRIX mode.
Solve your system now
Use MATRIX mode for the inverse method, or EQN mode to enter coefficients directly.
Open the equation solver →