Adjoint of a Matrix - Formula, Steps, Examples

What Is the Adjoint of a Matrix?

The adjoint of a square matrix $A$ — written $\operatorname{adj}(A)$ — is defined as the transpose of the matrix of cofactors.

Two preliminary definitions you need:

• Minor $M_{ij}$ of an entry: the determinant of the submatrix obtained by deleting row $i$ and column $j$.

• Cofactor $C_{ij}$ of an entry: $C_{ij} = (-1)^{i+j} M_{ij}$ — the minor with a checkerboard-pattern sign.

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

$$\operatorname{adj}(A) = (C_{ij})^T = (C_{ji})$$

The Connection to the Inverse

The adjoint exists for the sake of computing the inverse:

$$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$$

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

Adjoint of a 2×2 Matrix

For $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$:

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

The pattern: swap the diagonal, negate the off-diagonal. This is exactly what the inverse formula does (before dividing by the determinant).

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

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

Verify the inverse formula: $\det(A) = 12 - 10 = 2$, so $A^{-1} = \tfrac{1}{2} \begin{pmatrix} 4 & -5 \ -2 & 3 \end{pmatrix}$, which matches the standard $2 \times 2$ inverse formula.

Adjoint of a 3×3 Matrix

For a $3 \times 3$ matrix, the calculation is more involved but follows the same recipe.

For $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{pmatrix}$:

Compute the cofactor of each entry:

$$C_{ij} = (-1)^{i+j} \det\bigl[\text{2x2 submatrix obtained by deleting row } i, \text{column } j\bigr]$$

The cofactor matrix is $(C_{ij})_{3 \times 3}$. The adjoint is its transpose.

Step-by-step for 3×3.

1. For each entry, compute its $2 \times 2$ minor (determinant of the submatrix).

2. Apply the sign pattern $(-1)^{i+j}$ — checkerboard with $+$ in position $(1,1)$.

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 $A = \begin{pmatrix} 4 & 7 \ 2 & 6 \end{pmatrix}$.

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

Standard — 3×3 Adjoint

Find the adjoint of $A = \begin{pmatrix} 1 & 2 & 3 \ 0 & 1 & 4 \ 5 & 6 & 0 \end{pmatrix}$.

Compute cofactors (showing key ones):

• $C_{11} = +\det\begin{pmatrix} 1 & 4 \ 6 & 0 \end{pmatrix} = 0 - 24 = -24$

• $C_{12} = -\det\begin{pmatrix} 0 & 4 \ 5 & 0 \end{pmatrix} = -(0 - 20) = 20$

• $C_{13} = +\det\begin{pmatrix} 0 & 1 \ 5 & 6 \end{pmatrix} = 0 - 5 = -5$

• $C_{21} = -\det\begin{pmatrix} 2 & 3 \ 6 & 0 \end{pmatrix} = -(0 - 18) = 18$

• $C_{22} = +\det\begin{pmatrix} 1 & 3 \ 5 & 0 \end{pmatrix} = 0 - 15 = -15$

• $C_{23} = -\det\begin{pmatrix} 1 & 2 \ 5 & 6 \end{pmatrix} = -(6 - 10) = 4$

• $C_{31} = +\det\begin{pmatrix} 2 & 3 \ 1 & 4 \end{pmatrix} = 8 - 3 = 5$

• $C_{32} = -\det\begin{pmatrix} 1 & 3 \ 0 & 4 \end{pmatrix} = -(4 - 0) = -4$

• $C_{33} = +\det\begin{pmatrix} 1 & 2 \ 0 & 1 \end{pmatrix} = 1 - 0 = 1$

Cofactor matrix: $$C = \begin{pmatrix} -24 & 20 & -5 \ 18 & -15 & 4 \ 5 & -4 & 1 \end{pmatrix}$$

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

Stretch — Use Adjoint to Find Inverse

Using the 3×3 result above, find $A^{-1}$.

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

$$\det(A) = 1 \cdot C_{11} + 2 \cdot C_{12} + 3 \cdot C_{13}$$

Note: the original cofactors are signed already.

$$\det(A) = 1(-24) + 2(20) + 3(-5) = -24 + 40 - 15 = 1$$

So $A^{-1} = \tfrac{1}{1} \operatorname{adj}(A) = \operatorname{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \ 20 & -15 & -4 \ -5 & 4 & 1 \end{pmatrix}$.

Why Does the Adjoint Matter? (The Real-World GROUND)

"The adjoint is the bridge between the determinant and the inverse." — linear-algebra heuristic.

The adjoint is not the fastest practical way to compute an inverse for large matrices — Gaussian elimination is faster for matrices larger than $3 \times 3$. 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 $A\vec{x} = \vec{b}$, Cramer's rule expresses each unknown as a ratio of determinants — directly using cofactors and adjoint logic.

• Theoretical proofs. The adjoint is central to proving properties of determinants, eigenvalues, and the relationship between $A$ and $A^{-1}$.

• Symbolic algebra systems. Software like Mathematica and SymPy use adjoint-based formulas for symbolic matrix inversion where Gaussian elimination would produce messy intermediate expressions.

The adjoint formula for the inverse was first established by Pierre-Simon Laplace in the late 18th century, building on Gabriel Cramer's 1750 work on solving systems of equations (now known as Cramer's rule).

A Worked Example

Find the adjoint of $B = \begin{pmatrix} 1 & 3 \ 2 & 5 \end{pmatrix}$.

The intuitive (wrong) approach. A student computes the cofactor matrix and forgets to transpose:

$$\text{cofactor matrix} = \begin{pmatrix} 5 & -2 \ -3 & 1 \end{pmatrix}$$

The student writes this as the adjoint.

Why it fails. The adjoint is the transpose of the cofactor matrix — not the cofactor matrix itself. For $2 \times 2$ matrices, this means swapping the off-diagonal entries.

The correct method. Cofactor matrix above. Transpose:

$$\operatorname{adj}(B) = \begin{pmatrix} 5 & -3 \ -2 & 1 \end{pmatrix}$$

Verify: $\det(B) = 5 - 6 = -1$. $B^{-1} = \tfrac{1}{-1}\operatorname{adj}(B) = \begin{pmatrix} -5 & 3 \ 2 & -1 \end{pmatrix}$. Check $B \cdot B^{-1} = I$ ✓.

What Are the Most Common Mistakes With the Adjoint?

Mistake 1: Forgetting to transpose

The fix: The adjoint is the transpose of the cofactor matrix — always do the transpose step.

Mistake 2: Wrong sign pattern

Where it slips in: Putting $+$ in $(1,2)$ position. The pattern is $+, -, +, -, \ldots$ starting with $+$ in $(1,1)$.

The fix: Memorise the checkerboard: top-left is $+$, and signs alternate from there.

Mistake 3: Confusing minor with cofactor

Where it slips in: Forgetting the $(-1)^{i+j}$ factor.

The fix: Minor = determinant of submatrix (always positive when computed). Cofactor = minor with the sign applied.

Key Takeaways

• The adjoint $\operatorname{adj}(A)$ is the transpose of the cofactor matrix.

• Cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ — the minor with the checkerboard sign.

• Connection to the inverse: $A^{-1} = \dfrac{1}{\det(A)} \operatorname{adj}(A)$ (when $\det(A) \neq 0$).

• For 2×2: swap the diagonal, negate the off-diagonal — that's the adjoint.

• For 3×3 and larger: compute every cofactor, build the cofactor matrix, transpose.

A Practical Next Step

Try these three before moving on to determinant expansions and Cramer's rule.

1. Find the adjoint of $\begin{pmatrix} 6 & 2 \ 1 & 5 \end{pmatrix}$.

2. Use the adjoint formula to find the inverse of $\begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix}$.

3. Compute the cofactor $C_{23}$ of $\begin{pmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{pmatrix}$.