Skip to content
Hard

Solving Simultaneous Equations Using Matrices

Simple Explanation

A system of two linear equations can be written as a single matrix equation, AX = B, where A holds the coefficients, X holds the unknowns, and B holds the constants. Multiplying both sides by A⁻¹ solves the whole system at once: X = A⁻¹B.

Why Do We Need It?

This method solves an entire system of equations — potentially with many variables — using exactly the same two steps (find A⁻¹, then multiply) every time, rather than needing a different elimination or substitution strategy for each new system.

Formula

Inverse of a 2×2 Matrix

For A = [[a, b], [c, d]]: A⁻¹ = (1/det(A)) · [[d, −b], [−c, a]] (det(A) ≠ 0)

The inverse of a 2×2 matrix — swap the two diagonal entries, negate the two off-diagonal entries, then divide every entry by the determinant.

A⁻¹
— the inverse of A, satisfying A·A⁻¹ = A⁻¹·A = I (the identity matrix)
det(A)
— the determinant, ad − bc (must be nonzero)

When to use it: Whenever you need to "undo" a 2×2 matrix's effect, or solve a matrix equation of the form AX = B.

Worked Example

Solve a system of equations using matrices

Solve the system x + 2y = 4 and 3x + 5y = 11 using matrices.

    Why Does This Work?

    Multiplying both sides of AX=B by A⁻¹ on the left gives A⁻¹AX = A⁻¹B — and since A⁻¹A is always the identity matrix I (by definition of the inverse), and IX = X for any matrix X, this simplifies directly to X = A⁻¹B, isolating the unknowns in one step.

    Real-Life Example

    Balancing a mixture recipe

    A chemist needs to find the exact amounts of two solutions to mix to hit two separate target concentrations, described by two simultaneous equations.

    Writing the constraints as a matrix equation and solving with X=A⁻¹B gives the exact required amounts directly, especially useful when scaled up to many simultaneous constraints.

    Practice

    Solve 2x + y = 7 and x + 3y = 11 using matrices. Find x. (Hint: A=[[2,1],[1,3]], det=5.)

    Hard

    Common mistake

    Multiplying by A⁻¹ on the wrong side, or in the wrong order (BA⁻¹ instead of A⁻¹B) — matrix multiplication is not commutative, so A⁻¹ must be multiplied on the LEFT of both sides, consistently.

    Quick Review

    • Write the system as AX = B, then solve with X = A⁻¹B.
    • Works because A⁻¹A = I, the identity matrix, which leaves X unchanged.
    • Multiply A⁻¹ on the same side (left) of both A and B, consistently.
    ← Back to Matrices