Property A(adj A) = |A|I

Definitions. Let A = [aᵢⱼ] be of order n and Aᵢⱼ the cofactor of aᵢⱼ. Then adj A = [Aᵢⱼ]ᵀ, so the (i, j) entry of adj A is Aⱼᵢ.

Proof that A(adj A) = |A| I

The (i, j) entry of A(adj A) is

Σₖ aᵢₖ · Aⱼₖ      (k = 1 to n)

Two cases are used from the properties of determinants:

1. i = j: Σₖ aᵢₖ Aᵢₖ is the expansion of |A| along row i, so it equals |A|.
2. i ≠ j: Σₖ aᵢₖ Aⱼₖ is the expansion along row j of the determinant obtained by replacing row j with row i. That determinant has two identical rows, so its value is 0.

Hence all diagonal entries are |A| and all off-diagonal entries are 0:

A(adj A) = [|A|  0  …  0 ]
           [ 0  |A| …  0 ] = |A| Iₙ
           [ …          … ]
           [ 0   0  … |A|]

Proof that (adj A)A = |A| I

The (i, j) entry of (adj A)A is Σₖ Aₖᵢ aₖⱼ.

• i = j: this is the expansion of |A| along column i, so it equals |A|.
• i ≠ j: this is the expansion along column i of the determinant obtained by replacing column i with column j. That determinant has two identical columns, so it equals 0.

So (adj A)A = |A| Iₙ also.

Therefore A(adj A) = (adj A)A = |A| Iₙ. ∎

Consequence: if |A| ≠ 0, then A⁻¹ = adj A / |A|.