Adjoint of a Matrix: Properties
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Direct answer
The adjoint of a square matrix A is the transpose of its cofactor matrix, and one identity generates every exam property: A(adj A) = (adj A)A = |A|I. From it flow |adj A| = |A|^{n−1}, adj(adj A) = |A|^{n−2}A, (adj A)⁻¹ = adj(A⁻¹) = A/|A|, and the reversal law adj(AB) = adj(B)·adj(A). For order 2 there is a shortcut worth more than the definition: [[a, b], [c, d]] maps to [[d, −b], [−c, a]] — swap the diagonal, negate the off-diagonal, done.
What you must remember
- Master identity: A·adj A = adj A·A = |A|I; if A is singular the product collapses to the zero matrix.
- 2×2 shortcut: adj [[a, b], [c, d]] = [[d, −b], [−c, a]] — no cofactor machinery needed at order 2.
- Determinant rules: |adj A| = |A|^{n−1} for order n; so for |A| = 3 and n = 4, |adj A| = 27.
- Iterated adjoint: adj(adj A) = |A|^{n−2}A; at n = 2 this reduces to adj(adj A) = A — the adjoint is an involution on 2×2 matrices.
- Inverse links: (adj A)⁻¹ = adj(A⁻¹) = A/|A|; adj A is invertible exactly when A is.
- Reversal and scalars: adj(AB) = adj(B)·adj(A) — order flips, as with transposes and inverses; adj(kA) = k^{n−1}·adj A; adj(Aᵀ) = (adj A)ᵀ.
- Rank garnish (Advanced): for order n, rank(adj A) = n when rank(A) = n, equals 1 when rank(A) = n − 1, and adj A = 0 when rank(A) ≤ n − 2.
One matrix that verifies four properties
Let A = [[2, 3], [1, 4]] with |A| = 8 − 3 = 5. The shortcut gives adj A = [[4, −3], [−1, 2]]. Check the master identity: A(adj A) = [[2×4 − 3×1, −6 + 6], [4 − 4, −3 + 8]] = [[5, 0], [0, 5]] = 5I = |A|I. Check the determinant rule: |adj A| = 8 − 3 = 5 = |A|¹, as |A|^{n−1} demands at n = 2. Check the involution: adj(adj A) = adj [[4, −3], [−1, 2]] = [[2, 3], [1, 4]] = A, exactly what |A|^{n−2}A becomes at n = 2. One 2×2 matrix, four statements confirmed in four lines — reproducing this chain before the formulae-listing habit sets in fixes the exponents far better than rote learning. For contrast at order 3, if |A| = 2 then |adj A| = 4 and adj(adj A) = |A|A = 2A: the same identities, with the exponents now visibly doing work.
Where students slip
The reversal law is the top casualty: adj(AB) = adj(A)·adj(B) is false — adjoint reverses order exactly like transpose and inverse, and determinant-style habits (where order does not matter) pull students the wrong way. The exponents are the second: |adj A| = |A|^{n−1} and adj(adj A) = |A|^{n−2}A depend on the order n, and applying the n = 2 special case adj(adj A) = A to a 3×3 matrix is a reliable wrong answer. Third, singular behaviour: when |A| = 0 the adjoint still exists but A(adj A) = 0, so dividing by |A| to "find" an inverse is illegitimate — Advanced questions build directly on this edge. Finally, remember the adjoint is defined for square matrices only; a 2×3 matrix has minors and cofactors of entries but no adjoint.
Frequently asked questions
What is the adjoint of a 2×2 matrix?
Swap the diagonal entries, negate the off-diagonal entries: adj [[a, b], [c, d]] = [[d, −b], [−c, a]].
Why does adj(AB) reverse the order?
Taking cofactors of a product mixes entries from both factors, and the resulting algebra gives adj(AB) = adj(B)·adj(A) — the same reversal transposes and inverses obey.
What is |adj A| when |A| = 2 and n = 3?
|A|^{n−1} = 2² = 4; in general the determinant of the adjoint is the determinant raised to one less than the order.
Is the adjoint defined for rectangular matrices?
No — adjoint requires a full set of cofactors arranged square, so it exists only for square matrices.
What happens to adj A when A is singular?
The adjoint still exists and A(adj A) = 0; if rank(A) ≤ n − 2 the adjoint is the zero matrix, and if rank(A) = n − 1 it has rank 1.