Inverse Matrix Calculator

Invert square matrices with the 2×2 formula, Gauss-Jordan, or the adjugate method.

Inverse Matrix Calculator

Calculate matrix inverses with step-by-step solutions using multiple methods

Matrix Input

Matrix Properties

Size:2×2

About Matrix Inverses

Existence: A matrix has an inverse if and only if its determinant is non-zero (the matrix is non-singular).
Uniqueness: If a matrix inverse exists, it is unique. There is exactly one inverse matrix for each invertible matrix.
Properties: (AB)⁻¹ = B⁻¹A⁻¹ and (A⁻¹)⁻¹ = A
Applications: Solving linear systems, transformations, and many areas of mathematics and engineering.

What an inverse is

For a square matrix A, the inverse A⁻¹ satisfies A A⁻¹ = A⁻¹ A = I (identity). Only matrices with det(A) ≠ 0 have one. Those are non-singular; det = 0 means singular—no inverse.

Main use: Ax = b becomes x = A⁻¹b when A is invertible. In numerical work you often solve via LU/QR instead of forming A⁻¹ explicitly, but the concept still matters.

Methods

2×2 direct formula

A = [[a,b],[c,d]] → A⁻¹ = (1/(ad−bc)) [[d,−b],[−c,a]]

Fastest for 2×2 when the determinant is nonzero.

Gauss-Jordan

Augment [A | I], row-reduce to [I | A⁻¹]. Works for any size; same idea as solving systems by elimination.

Adjugate

A⁻¹ = (1/det(A)) adj(A)

Cofactors → transpose → scale. Useful for 3×3 by hand; expensive for larger n.

Useful identities

  • (AB)⁻¹ = B⁻¹ A⁻¹
  • (Aᵀ)⁻¹ = (A⁻¹)ᵀ
  • (A⁻¹)⁻¹ = A
  • det(A⁻¹) = 1/det(A)
  • Orthogonal: A⁻¹ = Aᵀ

Frequently Asked Questions

When is there no inverse?

When det(A) = 0—rows or columns linearly dependent. The map collapses dimension and cannot be undone uniquely.

Which method should I use?

2×2: direct formula. 3×3 by hand: adjugate or Gauss-Jordan. Larger or production code: factorization (LU with pivoting), not a hand adjugate.

How do I check the answer?

Multiply A × A⁻¹. You want I. Tiny off-diagonal junk (~1e−15) is floating-point noise.

Singular vs non-singular?

Non-singular: invertible, det ≠ 0. Singular: det = 0, no inverse, no unique solution for Ax = b in general.

Why avoid forming inverses in software?

Cost and stability. Solving Ax = b via decomposition is usually cheaper and better conditioned than computing A⁻¹ then multiplying.

Related tools

Related tools