Matrix Inversion: The Adjugate Matrix and Invertibility Criteria

A square matrix A is invertible if there is a matrix A⁻¹ with AA⁻¹ = A⁻¹A = I; this happens exactly when det A ≠ 0, and the inverse is given by A⁻¹ = adj(A)/det A, where the adjugate adj(A) is the transpose of the matrix of cofactors. In brief: for [[3, 1], [2, 5]], det = 13 and the inverse is (1/13)[[5, −1], [−2, 3]]. The invertibility criteria also include rank n, a reduced row echelon form equal to I, and the equation Ax = 0 having only the zero solution.

The inverse matrix solves Ax = b as x = A⁻¹b and undoes a linear transformation. This guide gives the 2 × 2 formula, the adjugate method for larger matrices, the list of equivalent invertibility criteria and the rules for products and transposes, with worked examples and a table.

What is the inverse of a 2 × 2 matrix?

For A = [[a, b], [c, d]] with det A = ad − bc ≠ 0, A⁻¹ = (1/(ad − bc)) [[d, −b], [−c, a]]: swap the diagonal entries and change the signs of the others.

Example 1: a 2 × 2 inverse

A = [[3, 1], [2, 5]], det = 13: A⁻¹ = (1/13) [[5, −1], [−2, 3]]. Check: [[3, 1], [2, 5]] · [[5, −1], [−2, 3]] = [[15 − 2, −3 + 3], [10 − 10, −2 + 15]] = 13 I ✓.

Example 2: a singular matrix

[[1, 2], [2, 4]] has det = 0, so it has no inverse. For which k is [[1, k], [k, 4]] invertible? det = 4 − k² ≠ 0, so k ≠ 2 and k ≠ −2.

How does the adjugate formula work?

The cofactor C_ij = (−1)^i+j M_ij is the minor M_ij (the determinant of A with row i and column j deleted) taken with the sign (−1)^i+j. The cofactor matrix C = (C_ij) transposed is the adjugate: adj(A) = C^T. It satisfies A · adj(A) = adj(A) · A = (det A) I, which follows from the Laplace expansion (and from the fact that expanding along the “wrong” row gives a determinant with two equal rows, hence 0). Therefore A⁻¹ = adj(A)/det A.

Example 3: a 3 × 3 inverse

A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]]. det A = 1(0 − 24) − 2(0 − 20) + 3(0 − 5) = −24 + 40 − 15 = 1. The cofactors:

• C₁₁ = det [[1, 4], [6, 0]] = −24, C₁₂ = −det [[0, 4], [5, 0]] = 20, C₁₃ = det [[0, 1], [5, 6]] = −5.
• C₂₁ = −det [[2, 3], [6, 0]] = 18, C₂₂ = det [[1, 3], [5, 0]] = −15, C₂₃ = −det [[1, 2], [5, 6]] = 4.
• C₃₁ = det [[2, 3], [1, 4]] = 5, C₃₂ = −det [[1, 3], [0, 4]] = −4, C₃₃ = det [[1, 2], [0, 1]] = 1.

The cofactor matrix is [[−24, 20, −5], [18, −15, 4], [5, −4, 1]] and its transpose gives A⁻¹ = [[−24, 18, 5], [20, −15, −4], [−5, 4, 1]], since det A = 1. Check the first row of A times the columns of A⁻¹: (−24 + 40 − 15, 18 − 30 + 12, 5 − 8 + 3) = (1, 0, 0) ✓.

What are the invertibility criteria?

For an n × n matrix A the following statements are equivalent:

• A is invertible.
• det A ≠ 0.
• The rank of A is n.
• The reduced row echelon form of A is I, equivalently A is a product of elementary matrices.
• The columns (and the rows) of A are linearly independent.
• Ax = 0 has only the solution x = 0.
• Ax = b has exactly one solution for every b.
• 0 is not an eigenvalue of A.

In a square matrix, a one-sided inverse is automatically two-sided: AB = I implies BA = I.

What are the rules for inverses?

(A⁻¹)⁻¹ = A; (AB)⁻¹ = B⁻¹A⁻¹ (reverse order, like taking off socks and shoes); (A^T)⁻¹ = (A⁻¹)^T; (kA)⁻¹ = A⁻¹/k; det(A⁻¹) = 1/det A. Special cases: the inverse of a diagonal matrix has the reciprocal entries; the inverse of an orthogonal matrix is its transpose.

Example 4: the inverse of a product

A = [[1, 1], [0, 1]], B = [[1, 0], [1, 1]]. AB = [[2, 1], [1, 1]] has det 1, so (AB)⁻¹ = [[1, −1], [−1, 2]]. And B⁻¹A⁻¹ = [[1, 0], [−1, 1]] · [[1, −1], [0, 1]] = [[1, −1], [−1, 2]] ✓. The other order, A⁻¹B⁻¹ = [[1, −1], [0, 1]] · [[1, 0], [−1, 1]] = [[2, −1], [−1, 1]], is not equal.

Example 5: a diagonal and an orthogonal matrix

diag(2, −5, 4)⁻¹ = diag(1/2, −1/5, 1/4). The rotation R = [[0, −1], [1, 0]] has R⁻¹ = R^T = [[0, 1], [−1, 0]], the rotation by −90°.

How do you solve a system with the inverse?

If A is invertible, Ax = b has the unique solution x = A⁻¹b. For a single system elimination is cheaper, but with the inverse available many right-hand sides are easy.

Example 6: solving with an inverse

[[3, 1], [2, 5]] x = (1, 2). x = (1/13) [[5, −1], [−2, 3]] (1, 2) = (1/13)(3, 4) = (3/13, 4/13). Check: 9/13 + 4/13 = 1 ✓ and 6/13 + 20/13 = 2 ✓.

Example 7: a matrix equation

Solve AX = B for X, where A = [[3, 1], [2, 5]] and B = [[1, 0], [0, 2]]. Then X = A⁻¹B = (1/13) [[5, −1], [−2, 3]] · [[1, 0], [0, 2]] = (1/13) [[5, −2], [−2, 6]]. Note that the inverse multiplies from the left, in the same order as A.

Table of inverse formulas

Where does the inverse come from?

The inverse of a matrix was introduced by Arthur Cayley in 1855–1858, who gave the formula in terms of determinants and cofactors for small matrices. Earlier, Cramer’s rule (1750) already solved systems by quotients of determinants. The general statement about equivalent invertibility conditions took its modern form with the abstract linear algebra of the twentieth century.

Which mistakes are common?

Forgetting to transpose the cofactor matrix

The adjugate is the transpose of the cofactor matrix. Skipping the transpose gives a wrong inverse for non-symmetric matrices.

Inverting entry by entry

The inverse of a matrix is not the matrix of reciprocals of its entries, except for diagonal matrices.

Wrong order in (AB)⁻¹

The inverse reverses the order: B⁻¹A⁻¹.

Inverting a non-square or singular matrix

Only square matrices with a nonzero determinant have inverses. Rectangular matrices can have one-sided inverses, studied through the rank.