Matrix Inverse

What an inverse means

A matrix A acts on vectors by left multiplication: x ↦ A·x. The inverse A⁻¹ is the matrix that reverses that action. If A rotates the plane by 30°, A⁻¹ rotates it back by −30°. If A doubles the x-component and halves the y-component, A⁻¹ halves the x-component and doubles the y-component.

The defining equation is A·A⁻¹ = A⁻¹·A = I, where I is the identity matrix (1s on the diagonal, 0s elsewhere). For this equation to make sense, A must be square. For an inverse to exist, A must be nonsingular — det A ≠ 0.

The geometric reason singular matrices have no inverse: if det A = 0, the transformation A collapses some direction to zero, so multiple input vectors map to the same output. There is no way to recover the input uniquely; no inverse function exists.

Worked example — 2×2 inverse

Take A = [[2, 1], [3, 4]]. Compute det A = 2·4 − 1·3 = 5.

The 2×2 inverse formula is A⁻¹ = (1/det A) · [[d, −b], [−c, a]] (swap diagonal, negate off-diagonal, divide by det):

A⁻¹ = (1/5) · [[4, −1], [−3, 2]] = [[4/5, −1/5], [−3/5, 2/5]].

Verify: A·A⁻¹ = [[2, 1], [3, 4]] · [[4/5, −1/5], [−3/5, 2/5]] = [[8/5 − 3/5, −2/5 + 2/5], [12/5 − 12/5, −3/5 + 8/5]] = [[1, 0], [0, 1]]. ✓

Worked example — 3×3 inverse via cofactor adjugate

Take A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].

Step 1 — det A. Expand along the first row:

det A = 1·(1·0 − 4·6) − 2·(0·0 − 4·5) + 3·(0·6 − 1·5) = 1·(−24) − 2·(−20) + 3·(−5) = −24 + 40 − 15 = 1.

Step 2 — cofactor matrix C. C_ij = (−1)^i+j·M_ij, where M_ij is the minor (determinant of the 2×2 submatrix from deleting row i, column j).

C = [[−24, 20, −5], [18, −15, 4], [5, −4, 1]]

Step 3 — adjugate. adj(A) = C^T (transpose of the cofactor matrix):

adj(A) = [[−24, 18, 5], [20, −15, −4], [−5, 4, 1]]

Step 4 — divide by det. det A = 1, so

A⁻¹ = adj(A) = [[−24, 18, 5], [20, −15, −4], [−5, 4, 1]].

Quick sanity check: row 1 of A times column 1 of A⁻¹ = 1·(−24) + 2·20 + 3·(−5) = −24 + 40 − 15 = 1. ✓

Methods of finding A⁻¹

	Cofactor adjugate	Gauss–Jordan elimination	LU decomposition
Idea	A⁻¹ = (1/det A) · adj(A)	Row-reduce [A | I] until A becomes I; the right block becomes A⁻¹	Factor A = LU; solve LU·X = I column by column
Best for	Symbolic 2×2 and 3×3 by hand	4×4 and larger by hand	Numerical computation, especially for many right-hand sides
Asymptotic cost	O(n!) recursive, O(n⁴) with caching	O(n³)	O(n³) factorisation + O(n²) per solve
Numerical stability	Poor — division by det amplifies errors	Decent with partial pivoting	Excellent with partial pivoting (PA = LU)
Closed-form output	Yes — explicit formula in entries	No — algorithmic	Yes (factors), but not a single closed form
Reusable for many right-hand sides?	Yes (compute A⁻¹ once)	Yes	Yes — most efficient: factor once, solve many
Detects singularity	det = 0 obviously	Pivot becomes zero	U has zero on diagonal

Practical advice: by hand on small symbolic matrices, use cofactors. Numerically, almost never compute A⁻¹ explicitly — solve the system Ax = b directly via LU, which is roughly twice as fast and far more stable.

Algebraic properties

• (A⁻¹)⁻¹ = A. Inverting twice is the identity operation.
• (AB)⁻¹ = B⁻¹A⁻¹. Reverse order — undo the most recent transformation first.
• (Aᵀ)⁻¹ = (A⁻¹)ᵀ. Transpose and inverse commute.
• (cA)⁻¹ = (1/c)·A⁻¹ for scalar c ≠ 0.
• det(A⁻¹) = 1 / det A. Inverse scales volumes by the reciprocal factor.
• Eigenvalues of A⁻¹ are reciprocals of eigenvalues of A. Same eigenvectors.
• A is orthogonal ⇔ A⁻¹ = Aᵀ. Hugely useful — orthogonal matrices invert by transposition.

Classical applications

• Solving linear systems. Ax = b has the formal solution x = A⁻¹b. In practice you do not compute A⁻¹; you solve directly.
• Linear regression — normal equations. The least-squares estimator is β̂ = (XᵀX)⁻¹·Xᵀy. Every regression textbook displays the inverse, but real software computes it via QR or SVD for stability.
• Change of basis. If P is the change-of-basis matrix from one coordinate system to another, the same linear map represented in the new basis is P⁻¹·A·P. Diagonalisation A = P·D·P⁻¹ is the canonical example.
• Computer graphics. Camera and model matrices are inverted to map between world, view, and screen coordinates. Orthogonal rotation matrices are inverted just by transposing.
• Cryptography (Hill cipher). Encrypts blocks of letters as vectors multiplied by an invertible matrix; decryption uses the inverse modulo 26.
• Markov chain analysis. Stationary distributions and fundamental matrices are computed by inverting (I − Q) where Q is the transient submatrix.

Common mistakes

• Writing (AB)⁻¹ = A⁻¹B⁻¹. Wrong order. The correct identity is (AB)⁻¹ = B⁻¹A⁻¹. Drop this and almost everything in derivations breaks.
• Trying to invert a non-square matrix. Only square matrices have classical inverses. For rectangular matrices, use the Moore–Penrose pseudoinverse A⁺.
• Computing A⁻¹ explicitly when you only need A⁻¹b. Slower and less stable than solving Ax = b directly.
• Forgetting to check det ≠ 0. If det A = 0, no inverse exists. Numerical software returns garbage or warnings rather than refusing outright.
• Using the cofactor formula for large matrices. O(n!) growth makes it impractical beyond about 5×5. Switch to Gauss–Jordan or LU.
• Confusing A⁻¹ with 1/A. 1/A is not a defined operation on matrices — there is no entry-wise division that makes algebraic sense. Always write A⁻¹.
• Mixing right and left inverses. For square matrices, left inverse = right inverse = the inverse. For rectangular matrices, one-sided inverses can exist without the other.