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# Adjoint of a Matrix - Formula, Steps, Examples

## TL;DR
The adjoint of a matrix AAA (also called adjugate) is the transpose of the cofactor matrix. It's the bridge that connects the matrix to its inverse:

A−1=1det⁡(A)adj⁡(A)A^{-1} = \dfrac{1}{\det(A)} \operatorname{adj}(A)

## What Is the Adjoint of a Matrix?
The **adjoint** of a square matrix AAA — written adj⁡(A)\operatorname{adj}(A) — is defined as the **transpose of the matrix of cofactors**.

Two preliminary definitions you need:

- **Minor** MijM_{ij} of an entry: the determinant of the submatrix obtained by deleting row iii and column jjj.
- **Cofactor** CijC_{ij} of an entry: Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}

The cofactor matrix is the matrix of all CijC_{ij}. The adjoint is its transpose:

adj⁡(A)=(Cij)T=(Cji)\operatorname{adj}(A) = (C_{ij})^T = (C_{ji})

## The Connection to the Inverse
The adjoint exists for the sake of computing the inverse:

A−1=1det⁡(A)adj⁡(A)

This works whenever det⁡(A)≠0\det(A) \neq 0. For singular matrices (det⁡=0), the adjoint still exists, but division by zero makes the inverse undefined.

## Adjoint of a 2×2 Matrix
For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}:

adj⁡(A)=(d−b−ca)\operatorname{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

**Worked example.** Find the adjoint of A=(3524)A = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix}.

adj⁡(A)=(4−5−23)\operatorname{adj}(A) = \begin{pmatrix} 4 & -5 \\ -2 & 3 \end{pmatrix}

## Adjoint of a 3×3 Matrix
For A=(a11a12a13a21a22a23a31a32a33)A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}:

Compute the cofactor of each entry:

Cij=(−1)i+jdet⁡[2x2 submatrix obtained by deleting row i,column j]

**Step-by-step for 3×3.**
1. For each entry, compute its 2×2 minor.
2. Apply the sign pattern (−1)i+j.
3. Form the cofactor matrix.
4. Transpose: swap entries across the main diagonal.

## Three Worked Examples — Quick, Standard, Stretch

### Quick — 2×2 Adjoint
Find the adjoint of A=(4726)A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}.

adj⁡(A)=(6−7−24)\operatorname{adj}(A) = \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}

### Standard — 3×3 Adjoint
Find the adjoint of A=(123014560)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}.

Cofactor matrix: C=(−2420−518−1545−41)

Adjoint = transpose: adj⁡(A)=(−2418520−15−4−541)\operatorname{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}

### Stretch — Use Adjoint to Find Inverse
Using the 3×3 result above, find A−1.

First compute det⁡(A)\det(A) by expanding along the first row:

det⁡(A)=1⋅C11+2⋅C12+3⋅C13

det(A)=1(−24)+2(20)+3(−5)=−24+40−15=1

So A−1=11adj⁡(A)=adj⁡(A)=\begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}.

## Why Does the Adjoint Matter? 
The adjoint is not the fastest practical way to compute an inverse for large matrices — Gaussian elimination is faster for matrices larger than 3×3. But for small matrices and for theoretical work, the adjoint is essential:
- **Closed-form inverse for 2×2 and 3×3.** Matrices that appear in computer graphics rotations, physics linear transformations, and statistics covariance matrices are often this size, and the adjoint gives a clean formula.
- **Cramer's rule.** For solving Ax⃗=b⃗, Cramer's rule expresses each unknown as a ratio of determinants — directly using cofactors and adjoint logic.

## Key Takeaways
- **The adjoint** adj⁡(A) is the **transpose of the cofactor matrix**.
- **Cofactor Cij=(−1)i+jMij** — the minor with the checkerboard sign.
- **Connection to the inverse**: A−1=1det⁡(A)adj⁡(A) (when det⁡(A)≠0).

## A Practical Next Step
Try these three before moving on to determinant expansions and Cramer's rule.
1. Find the adjoint of (6215).
2. Use the adjoint formula to find the inverse of (2143).
3. Compute the cofactor C23 of (123456789).