Matrix Inverse Calculator

How to Use This Calculator

1. Choose 2x2 or 3x3, enter matrix entries — inverse appears instantly.
2. Use Verify tab to confirm A * A-inverse = I.
3. Use Professional to solve Ax=b and compute the condition number.

Formula

Example

Frequently Asked Questions

• How do you find the inverse of a 2x2 matrix? ▼
  For a 2×2 matrix A = [[a, b], [c, d]], the inverse is A⁻¹ = (1/det(A)) × [[d, −b], [−c, a]], where det(A) = ad − bc. The formula swaps the main diagonal entries, negates the off-diagonal entries, and divides everything by the determinant. Example: A = [[4, 7], [2, 6]]. det = 4·6 − 7·2 = 24 − 14 = 10. A⁻¹ = (1/10) × [[6, −7], [−2, 4]] = [[0.6, −0.7], [−0.2, 0.4]]. Verification: A × A⁻¹ should equal the identity matrix I. Common pitfall: forgetting to divide by det(A) — multiplying by the adjugate without scaling gives det(A) × I, not I.
• When does a matrix not have an inverse? ▼
  A square matrix A does not have an inverse — it is called singular — when its determinant equals zero. This happens when one or more rows (or columns) are linearly dependent, meaning at least one row can be expressed as a linear combination of the others. For a 2×2 matrix, det = 0 means one row is a scalar multiple of the other. Example: A = [[2, 4], [1, 2]] has det = 4 − 4 = 0 and is not invertible because row 1 = 2 × row 2. Geometrically, a singular matrix collapses the input space — vectors get mapped onto a line (2D) or plane (3D), so the transformation cannot be undone. Also note that non-square matrices (more rows than columns, or vice versa) do not have standard inverses.
• What is the condition number? ▼
  The condition number k(A) = ||A|| × ||A⁻¹|| measures how sensitive the solution x = A⁻¹b is to small changes in b or A. A condition number near 1 indicates a well-conditioned matrix — small errors in the input lead to small errors in the solution. A large condition number (e.g., 10⁶) means the matrix is ill-conditioned — tiny input perturbations cause large changes in the output. Example: if k(A) = 1000 and b has 3 digits of precision, the solution x may have only about 0 reliable digits. Common pitfall: a matrix can have a large condition number even if its determinant is not close to zero — det measures singularity while condition number measures numerical stability. Use the condition number, not the determinant, to diagnose numerical problems.
• What is the Moore-Penrose pseudoinverse? ▼
  The Moore-Penrose pseudoinverse A⁺ generalizes the matrix inverse to non-square or singular matrices. For a full-rank matrix A with more rows than columns, A⁺ = (AᵀA)⁻¹Aᵀ gives the least-squares solution to Ax ≈ b — minimizing ||Ax − b||². For a full-rank matrix with more columns than rows, A⁺ = Aᵀ(AAᵀ)⁻¹ gives the minimum-norm solution. For square invertible matrices, A⁺ = A⁻¹ exactly. Example: for a tall 3×2 matrix with full column rank, A⁺ finds the best-fit x. Pitfall: if A is rank-deficient (some columns are linearly dependent), computing (AᵀA)⁻¹ fails because AᵀA is singular — you need SVD-based pseudoinverse instead.
• How do I solve Ax=b using the inverse? ▼
  When A is invertible, the linear system Ax = b has the unique solution x = A⁻¹b. You compute A⁻¹ once and then multiply it by b to get x. Example: A = [[4, 7], [2, 6]], b = [1, 0]. A⁻¹ = [[0.6, −0.7], [−0.2, 0.4]]. x = [[0.6, −0.7], [−0.2, 0.4]] × [1, 0] = [0.6, −0.2]. Check: 4(0.6) + 7(−0.2) = 2.4 − 1.4 = 1 ✓ and 2(0.6) + 6(−0.2) = 1.2 − 1.2 = 0 ✓. A common pitfall is using this method when A is ill-conditioned — the computed A⁻¹ accumulates rounding errors and x may be inaccurate. For large systems or repeated solves, Gaussian elimination (LU decomposition) is numerically preferred over explicitly computing A⁻¹.

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