Adjoint of a Matrix
2026-02-28 01:31 Diff
https://brightchamps.com/en-us/math/algebra/adjoint-of-a-matrix

To find the adjoint of a matrix, follow the steps mentioned below:

1. Compute the Minor of each element:

For each component​ aij in the matrix, remove the i-th row and j-th column to form a sub-matrix. Compute the determinant of this sub-matrix; this value is the minor mij of the element ​aij

For example, a11 = 1

Now we need to remove the first row and first column

\( \begin{bmatrix} {4} & {5} \\ {0} &{6}  \end{bmatrix}\)


The determinant will be
det = (4)(6) − (5)(0) = 24
M11 = 24

2. Determine the Cofactor: The cofactor cij ​is given by

\(c_{ij} = (-1)^{i+j}. m _ {ij}\)

This accounts for the sign change based on the position of the element.

For example, a11 = 1
C11 = (−1)1 + 1 × 24 = 24

3. Form the Cofactor matrix: Arrange all the cofactors cij into a matrix.

Cofactor matrix = \( \begin{bmatrix} {24} & {-20} &{5}\\ {-5} &{6} &{-1}\\{-4} &{4} &{4} \end{bmatrix}\)

4. Transpose the Cofactor matrix: Take the transpose of the cofactor matrix to obtain the adjoint matrix.

Adj(A) = \( \begin{bmatrix} {24} & {-5} &{-4}\\ {-20} &{6} &{4}\\{-5} &{-1} &{4} \end{bmatrix}\)


Adjoint of a 2 × 2 matrix

The adjoint of a square matrix is gained by first calculating its cofactor matrix and then taking the transpose of that matrix. This adjoint is specifically useful in computing the inverse of the original matrix, provided its determinant is non-zero.

Steps to Find the Adjoint of a 2 × 2 matrix

The matrix given is:

\(A= \begin{bmatrix} {a} & {b} \\{c} &{d}  \end{bmatrix}\)

  1. Compute the Cofactor matrix: For each element, calculate its cofactor:
     

    c11 = d 

    c12 = -c

    c21 = -b

    c22 = a


    So, the cofactor matrix is:
     

    \(C= \begin{bmatrix} {d} & {-c} \\{-b} &{a}  \end{bmatrix}\)
     

  2. Transpose the Cofactor matrix:

    The adjoint is the transpose of the cofactor matrix:

    adj(A) = CT =  \(C= \begin{bmatrix} {d} & {-b} \\{-c} &{a}  \end{bmatrix}\)

    Example
    Given the matrix:

    \(C= \begin{bmatrix} {3} & {-6} \\{-4} &{8}  \end{bmatrix}\)

  • 1. Cofactor matrix:

    \(C= \begin{bmatrix} {8} & {4} \\{-6} &{3}  \end{bmatrix}\)

  •  2. Adjoint:

    adj(A) = CT = \(\begin{bmatrix} {8} & {-6} \\{4} &{3}  \end{bmatrix}\)


Adjoint of 3 × 3 matrices


Every adjoint of a square matrix is the transpose of its cofactor matrix. It is essential for calculating the inverse of a matrix, provided the determinant is non-zero.

Step-by-Step Process

The matrix given is:

\(A = \begin{bmatrix}{a_{11}} & {a_{12} }&{a_{13}} \\ a_{21} & a_{22} &a_{23} \\a_{31} & a_{32} &a_{33} \end{bmatrix}\)

1. Compute the minors of all elements:
Calculate mij of all elements by deleting the i-th row and j-th column of A, and finding the determinant of the sub-matrix.

2. Compute the cofactors of all elements:

For each element, aij​ calculate its cofactor cij using the formula:
\(c_{ij}=(-1)^{i+j} . det(m^{ij})\)

3. Form the Cofactor matrix:
Arrange all the cofactors cij​ into a matrix.


4. Transpose the Cofactor matrix:

 The adjoint (adjoint) is the transpose of the cofactor matrix:
 adj(A) = CT

The following chart gives the steps in summarized form for calculating adjoint of a matrix:

Practice Problem:
Given the matrix:

\(A = \begin{bmatrix}{1} & {2}&{3} \\ 4 & 5&6 \\7 & 8 &9\end{bmatrix}\)

1. Compute the minors:

\(M_{11} = 5 \times 9 - 6\times 8 = 45 - 48 = -3\\ M_{12} = 4 \times 9 - 6\times 7 = 36 - 42 = -6\\ M_{13} = 4 \times 8 - 5 \times 7 = 32 - 35 = -3 \\ M_{21} = 2 \times 9 = 3 \times 8 = 18 - 24 = -6\\ M_{22} = 1 \times 9 - 3 \times 7 = 9 - 21 = -12\\ M_{23} = 1 \times 8 - 2\times 7 = 8 - 14 = 6\\ M_{31} = 2 \times 6 - 5 \times 3 = 12 - 15 = -3\\ M_{32} = 1 \times6 - 4 \times 3 = 6 - 12 = -6\\ M_{33} = 1 \times 5 - 4 \times 2 = 5 - 8 = -3\)

2. Calculate the cofactors off all elements:

\(C_{11} -3\\ C_{12} = 6\\ C_{13} = -3 \\ C_{21} = 6\\ C_{22} = -12\\ C_{23} = - 6\\ C_{31} = -3\\ C_{32} = 6\\ C_{33} = -3\)

3. Form the Cofactor matrix:


\(A = \begin{bmatrix}{-3} & {6}&{-3} \\ 6 & -12&-6 \\-3 & 6 &-3\end{bmatrix}\)


4. Transpose the Cofactor matrix:


\(A = \begin{bmatrix}{-3} & {6}&{-3} \\ 6 & -12&6 \\-3 & -6 &-3\end{bmatrix}\)