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Adjugate

Adjugate
Classical Adjoint

The matrix formed by taking the transpose of the cofactor matrix of a given original matrix. The adjugate of matrix A is often written adj A.

Note: In the past, the term for adjugate used to be adjoint. The name has changed to avoid ambiguity with a different defintition of the term adjoint. Sometimes the adjugate is called the classical adjoint to refer to this older term. Consult a book on linear algebra for more information.

 

Example:

Find the adjugate of the following matrix:

\[{\rm{A}} = \left[ {\begin{array}{*{20}c} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \\\end{array}} \right]\]
Solution:

First find the cofactor of each element.

\[\begin{array}{*{20}{l}}{{{\rm{A}}_{11}} = \left| {\begin{array}{*{20}{c}}4&5\\0&6\end{array}} \right| = 24}&{{{\rm{A}}_{12}} = - \left| {\begin{array}{*{20}{c}}0&5\\1&6\end{array}} \right| = 5}&{{{\rm{A}}_{13}} = \left| {\begin{array}{*{20}{c}}0&4\\1&0\end{array}} \right| = - 4}\\{}&{}&{}\\{{{\rm{A}}_{21}} = - \left| {\begin{array}{*{20}{c}}2&3\\0&6\end{array}} \right| = - 12}&{{{\rm{A}}_{22}} = \left| {\begin{array}{*{20}{c}}1&3\\1&6\end{array}} \right| = 3}&{{{\rm{A}}_{23}} = - \left| {\begin{array}{*{20}{c}}1&2\\1&0\end{array}} \right| = 2}\\{}&{}&{}\\{{{\rm{A}}_{31}} = \left| {\begin{array}{*{20}{c}}2&3\\4&5\end{array}} \right| = - 2}&{{{\rm{A}}_{32}} = - \left| {\begin{array}{*{20}{c}}1&3\\0&5\end{array}} \right| = - 5}&{{{\rm{A}}_{33}} = \left| {\begin{array}{*{20}{c}}1&2\\0&4\end{array}} \right| = 4}\end{array}\]

As a result the cofactor matrix of A is

\[\left[ {\begin{array}{*{20}{c}}{24}&5&{ - 4}\\{ - 12}&3&2\\{ - 2}&{ - 5}&4\end{array}} \right]\]

Finally the adjugate of A is the transpose of the cofactor matrix:

\[\left[ {\begin{array}{*{20}{c}}{24}&{ - 12}&{ - 2}\\5&3&{ - 5}\\{ - 4}&2&4\end{array}} \right]\]

 

See also

Inverse of a matrix

Key Formula

A−1=1det⁡(A)adj⁡(A)A^{-1} = \frac{1}{\det(A)}\operatorname{adj}(A)
Where:
  • AA = An n × n square matrix that is invertible (det(A) ≠ 0)
  • adj⁡(A)\operatorname{adj}(A) = The adjugate of A, formed by transposing the cofactor matrix of A
  • det⁡(A)\det(A) = The determinant of A
  • A−1A^{-1} = The inverse of A

Worked Example

Problem: Find the adjugate of the 3×3 matrix B and use it to compute B⁻¹. B = [[2, 1, 0], [0, 3, 1], [1, 0, 2]]
Step 1: Compute the cofactors for row 1. For each entry, delete the row and column containing it, compute the 2×2 determinant, and apply the checkerboard sign pattern.
C11=+∣3102∣=6,C12=−∣0112∣=1,C13=+∣0310∣=−3C_{11} = +\begin{vmatrix}3&1\\0&2\end{vmatrix} = 6, \quad C_{12} = -\begin{vmatrix}0&1\\1&2\end{vmatrix} = 1, \quad C_{13} = +\begin{vmatrix}0&3\\1&0\end{vmatrix} = -3
Step 2: Compute the cofactors for row 2.
C21=−∣1002∣=−2,C22=+∣2012∣=4,C23=−∣2110∣=1C_{21} = -\begin{vmatrix}1&0\\0&2\end{vmatrix} = -2, \quad C_{22} = +\begin{vmatrix}2&0\\1&2\end{vmatrix} = 4, \quad C_{23} = -\begin{vmatrix}2&1\\1&0\end{vmatrix} = 1
Step 3: Compute the cofactors for row 3.
C31=+∣1031∣=1,C32=−∣2001∣=−2,C33=+∣2103∣=6C_{31} = +\begin{vmatrix}1&0\\3&1\end{vmatrix} = 1, \quad C_{32} = -\begin{vmatrix}2&0\\0&1\end{vmatrix} = -2, \quad C_{33} = +\begin{vmatrix}2&1\\0&3\end{vmatrix} = 6
Step 4: Form the cofactor matrix, then transpose it to get the adjugate.
Cofactor matrix=[61−3−2411−26],adj⁡(B)=[6−2114−2−316]\text{Cofactor matrix} = \begin{bmatrix}6&1&-3\\-2&4&1\\1&-2&6\end{bmatrix}, \quad \operatorname{adj}(B) = \begin{bmatrix}6&-2&1\\1&4&-2\\-3&1&6\end{bmatrix}
Step 5: Compute det(B) using cofactor expansion along row 1, then find the inverse.
det⁡(B)=2(6)+1(1)+0(−3)=13,B−1=113[6−2114−2−316]\det(B) = 2(6) + 1(1) + 0(-3) = 13, \quad B^{-1} = \frac{1}{13}\begin{bmatrix}6&-2&1\\1&4&-2\\-3&1&6\end{bmatrix}
Answer: The adjugate of B is [[6, −2, 1], [1, 4, −2], [−3, 1, 6]], and B⁻¹ = (1/13) · adj(B).

Another Example

This example shows the simpler 2×2 case, where a direct shortcut formula replaces the full cofactor-and-transpose procedure needed for larger matrices.

Problem: Find the adjugate of the 2×2 matrix C = [[5, 3], [2, 4]].
Step 1: For a 2×2 matrix, there is a shortcut. Given a general matrix [[a, b], [c, d]], the adjugate swaps the diagonal entries and negates the off-diagonal entries.
If C=[abcd], then adj⁡(C)=[d−b−ca]\text{If } C = \begin{bmatrix}a&b\\c&d\end{bmatrix}, \text{ then } \operatorname{adj}(C) = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}
Step 2: Apply this directly to our matrix.
adj⁡(C)=[4−3−25]\operatorname{adj}(C) = \begin{bmatrix}4&-3\\-2&5\end{bmatrix}
Step 3: Verify using the inverse formula. The determinant is det(C) = 5·4 − 3·2 = 14.
C−1=114[4−3−25]C^{-1} = \frac{1}{14}\begin{bmatrix}4&-3\\-2&5\end{bmatrix}
Answer: adj(C) = [[4, −3], [−2, 5]].

Frequently Asked Questions

What is the difference between the adjugate and the adjoint of a matrix?
The adjugate (also called the classical adjoint) is the transpose of the cofactor matrix and is used to compute a matrix inverse via determinants. The adjoint (in modern usage) typically refers to the conjugate transpose of a matrix, written A*, which is important in inner product spaces and functional analysis. Historically the word 'adjoint' was used for both concepts, which is why the name 'adjugate' was introduced to avoid confusion.
How do you use the adjugate to find the inverse of a matrix?
Divide the adjugate by the determinant: A⁻¹ = (1/det(A)) · adj(A). This works for any invertible square matrix. If det(A) = 0, the matrix is singular and has no inverse, so the formula does not apply.
Does the adjugate exist for singular (non-invertible) matrices?
Yes. The adjugate is defined for any square matrix, whether invertible or not. You can always compute the cofactor matrix and transpose it. However, for a singular matrix, you cannot use the adjugate to find an inverse because the determinant is zero. The identity A · adj(A) = det(A) · I still holds — both sides simply equal the zero matrix when det(A) = 0.

Adjugate (Classical Adjoint) vs. Adjoint (Conjugate Transpose)

Adjugate (Classical Adjoint)Adjoint (Conjugate Transpose)
DefinitionTranspose of the cofactor matrixTranspose of the complex conjugate of a matrix (A*)
Notationadj(A)A* or A†
Key formulaA⁻¹ = (1/det A) adj(A)For real matrices, A* = Aᵀ
When to useFinding matrix inverses via determinants, Cramer's rule proofsInner product spaces, unitary/Hermitian matrices, quantum mechanics
RequiresSquare matrix (any size)Any matrix (not necessarily square)

Why It Matters

The adjugate appears in the derivation of Cramer's rule and in the general determinant formula for matrix inverses, both of which are central topics in a linear algebra course. Understanding it solidifies your grasp of cofactors, determinants, and transposes — three concepts that work together frequently. In more advanced courses, the adjugate also plays a role in the Cayley-Hamilton theorem and in studying how matrices behave over rings (not just fields).

Common Mistakes

Mistake: Forgetting to transpose the cofactor matrix.
Correction: The adjugate is not the cofactor matrix itself — it is the transpose of the cofactor matrix. After computing all the cofactors and arranging them in a matrix, you must swap rows and columns to get adj(A).
Mistake: Getting the sign pattern wrong when computing cofactors.
Correction: The cofactor C_{ij} includes the sign factor (−1)^{i+j} multiplied by the minor M_{ij}. A common error is to forget this alternating sign or to apply it inconsistently. Use the checkerboard pattern: + − + / − + − / + − + for a 3×3 matrix.

Related Terms

  • Cofactor Matrix — The adjugate is its transpose
  • Cofactor — Individual signed minors that form the cofactor matrix
  • Transpose of a Matrix — Transposing the cofactor matrix yields the adjugate
  • Inverse of a Matrix — Computed using the adjugate divided by the determinant
  • Matrix — The adjugate is defined for square matrices
  • Determinant — Used with the adjugate to find the inverse
  • Minor — Determinant of the submatrix used to compute each cofactor