Compute the cofactor matrix and adjoint (adjugate) of any square matrix — with every minor determinant expanded step by step.
Matrix Setup
Results at a Glance
Step 1 — Sign Pattern
Step 2 — Cofactor Matrix C
Each entry $C_{ij} = (-1)^{i+j} \cdot M_{ij}$ where $M_{ij}$ is the determinant of the submatrix formed by deleting row $i$ and column $j$.
Click any cell in the cofactor matrix below to see its detailed calculation.
Detailed Calculations for Each Cofactor
Step 3 — Adjoint Matrix = Cᵀ
The adjoint (adjugate) of $A$ is the transpose of the cofactor matrix:
$$\text{adj}(A) = C^T \quad \Longrightarrow \quad [\text{adj}(A)]_{ij} = C_{ji}$$
Bonus — Inverse via Adjoint
Theory & Key Formulas
Minor $M_{ij}$ — The determinant of the $(n{-}1)\times(n{-}1)$ submatrix obtained by deleting row $i$ and column $j$ from $A$.
Cofactor $C_{ij}$ — The signed minor:
$$C_{ij} = (-1)^{i+j} M_{ij}$$
The sign pattern follows the checkerboard: $+$ when $i+j$ is even, $-$ when $i+j$ is odd.
Adjoint (Adjugate) — Transpose of the cofactor matrix:
$$\text{adj}(A) = C^T, \quad [\text{adj}(A)]_{ij} = C_{ji}$$
The cofactor of entry a(i,j) is C(i,j) = (−1)^(i+j) · M(i,j), where M(i,j) is the minor. The cofactor matrix collects all C(i,j), and the adjoint (adjugate) is its transpose. These satisfy A · adj(A) = det(A) · I, so for an invertible matrix the inverse is A⁻¹ = adj(A) / det(A).
Common questions
Is the adjoint the same as the cofactor matrix?
No. The adjoint is the transpose of the cofactor matrix.
How is the adjoint used to find an inverse?
Divide the adjoint by the determinant: A⁻¹ = adj(A) / det(A), valid when det(A) ≠ 0.