Quick answer

For square A, A · adj(A) = det(A) · I. Cofactor expansion gives det(A) as a signed sum along any row or column.

Formula

  • A · adj(A) = det(A) · I
  • det(A) = Σ_j a_ij C_ij along row i
  • Invertible iff det(A) ≠ 0

Introduction

Adjoints and determinants are paired topics in linear algebra courses. You rarely study one for long without the other.

Review adjoint matrix formula for cofactor notation, then use the Adjoint Matrix Calculator for adj(A) while you practice det(A) by expansion on the same matrix.

For inverse steps after the product check passes, see adjoint matrix and inverse matrix and divide adj(A) by det(A) only when det(A) ≠ 0.

Determinant role

det(A) is a scalar summary of A. For 2×2, det(A) = ad − bc. For larger sizes, use cofactor expansion or row operations as taught.

When det(A) = 0, rows or columns are linearly dependent in square cases, and A⁻¹ does not exist over the reals.

Multiplying A by adj(A) always yields det(A) times I, which is a strong check after you compute both objects.

Cofactor signs used in determinants match signs in adj(A). Reuse one row of cofactors from your adjoint work to find det(A) without starting over.

Pick one row with many zeros for fastest expansion. Zero entries kill corresponding terms in the sum.

Linked formulas

  • A · adj(A) = adj(A) · A = det(A) · I
  • det(A) = a_i1 C_i1 + ··· + a_in C_in (row i expansion)
  • A⁻¹ = adj(A) / det(A) when det(A) ≠ 0

If you already built the cofactor matrix for adj(A), reuse that row cofactors to find det(A).

Compare det(A) from expansion with det(A) from row echelon form if your course teaches both.

Can det(A) be zero while adj(A) is nonzero? Yes. Only the inverse formula fails.

Why multiply by I? The product must be a scalar multiple of the identity, which is how det(A) enters cleanly.

Determinant connections

  1. Choose an expansion row or column. Prefer a line with zeros to reduce arithmetic.
  2. Compute cofactors on that line. Use the same C_ij signs as for the full adjoint.
  3. Sum products a_ij C_ij. The sum equals det(A).
  4. Multiply A · adj(A). Confirm you receive det(A) · I. Fix cofactors first if the product fails.

Determinant and adjoint check

Let A = [[2, 0], [7, −3]]. Then det(A) = −6 and adj(A) = [[−3, 0], [−7, 2]] from the 2×2 shortcut.

Multiply to see A · adj(A) = −6 · I, matching det(A) = −6.

For 3×3, pick row 1 expansion if convenient, then still verify with the full product identity when time allows.

Practice finding det(A) and adj(A) for the same matrix before you attempt A⁻¹.