Adjoint of a Matrix - Formula, Steps, Examples

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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:

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:

Key Takeaways

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).