Singular matrices and the inverse

A square matrix is singular if |A| = 0 and non-singular if |A| ≠ 0. [[1, 2], [4, 8]] is singular; [[1, 2], [3, 4]] is non-singular, with determinant 4 − 6 = −2.

Product rule: |AB| = |A||B| for square matrices of the same order. So if A and B are both non-singular, so are AB and BA.

Taking determinants of (adj A)A = |A| I gives |adj A||A| = |A|ⁿ, so |adj A| = |A|ⁿ⁻¹. For order 3 this is |A|².

A square matrix is invertible if and only if it is non-singular. If AB = I then |A||B| = 1, so |A| ≠ 0; conversely, if |A| ≠ 0, dividing A(adj A) = |A| I by |A| shows that (1/|A|) adj A works as the inverse.

A⁻¹ = (1/|A|) adj A. For A with rows (1, 3, 3), (1, 4, 3), (1, 3, 4), |A| = 7 − 3 − 3 = 1, adj A has rows (7, −3, −3), (−1, 1, 0), (−1, 0, 1), and A⁻¹ equals adj A.

Worked example: A = [[2, 3], [1, −4]] and B = [[1, −2], [−1, 3]] give AB = [[−1, 5], [5, −14]] with |AB| = −11, and (AB)⁻¹ = (1/11)[[14, 5], [5, 1]], which matches B⁻¹A⁻¹.

An inverse can come from a matrix equation. A = [[2, 3], [1, 2]] satisfies A² − 4A + I = O. Multiplying by A⁻¹ gives A − 4I + A⁻¹ = O, so A⁻¹ = 4I − A = [[2, −3], [−1, 2]].

Because |A||A⁻¹| = |I| = 1, det(A⁻¹) = 1/det A.