Adjoint of a Matrix

Matrices are fundamental in mathematics and are crucial in various branches such as linear algebra, computer science, and physics. One important concept derived from matrices is the determinant, which is closely related to the adjoint of a matrix.

The adjoint (or adjugate) of a matrix is the transpose matrix of the cofactor of the given matrix. For any square matrix A to calculate its adjoint matrix we have to first calculate the cofactor matrix of the given matrix and then find its determinant.

To calculate the adjoint of a matrix, follow the following steps:

Step 1: Calculate the Minor of all the elements of the given matrix A.
Step 2: Find the Cofactor matrix C using the minor elements.
Step 3: Find the Adjoint matrix of A by taking the transpose of the cofactor matrix C.

For any 2×2 matrix A, the image of its Adjoint is shown below,

Minor of a Matrix

The minor of an element a_ij​ in a matrix is calculated by removing the row and column containing that element, and then finding the determinant of the remaining submatrix.

Steps to Find the Minor:

Step 1: Select the element a_ij whose minor you want to find.
Step 2: Remove the row and column containing a_ij.
Step 3: Calculate the determinant of the resulting submatrix. This is the minor M_ij.

For a 2×2 matrix, the minor is simply the determinant of the 1×1 matrix that remains after removing the element’s row and column.

Learn more about Minors and Cofactors.,

Cofactor of a Matrix

The Cofactor is the number we get when we remove the column and row of a designated element in a matrix. It means to take one element from a matrix and delete the entire row and column of that element from the matrix; which elements present in that matrix that is called the cofactor.

How to Find the Cofactor of a Matrix

To find the cofactor of an element of a matrix, we can use the following steps:

Step 1: Delete the entire row and column that contains element under consideration. 
Step 2: Take the remaining elements as it is in the matrix after Step 1.
Step 3: Find the determinant of the matrix formed in Step 2 which is called the minor of the element.
Step 4: Now use the formula for the cofactor of element a_ij i.e., (-1)^i+j M_ij where Mij is the minor of the element in the i^th row and j^th column which is already calculated in Step 3.
Step 5: Result of Step 4 is the cofactor of the element under consideration, and similarly, we can calculate the cofactor of each element of the matrix to find the cofactor matrix of the given matrix.

Example: Find the Cofactor Matrix of[Tex]\bold{A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}}        [/Tex].

Solution:

Given matrix is [Tex]A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}[/Tex]

Let’s find the cofactor of element in first row third column i.e., 3.

Step 1: Delete the entire row and column that contains element under consideration. 
i.e.,[Tex] \begin{bmatrix} \sout{1} & \sout{2} & \sout{3}\\ 7 & 4 & \sout{5} \\ 6 & 8 & \sout{9} \end{bmatrix}[/Tex]

Step 2: Take the remaining elements as it is in the matrix after Step 1.
i.e., [Tex]\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}[/Tex]

Step 3: Find the determinant of the matrix formed in Step 2 which is called the minor of the element.
Minor of 3 in [Tex]A  = \begin{vmatrix} 7 & 4 \\ 6 & 8 \end{vmatrix} = 56 – 24 = 32[/Tex]

Step 4: Now use the formula for the cofactor of element a_ij i.e., (-1)^i+j M_ij 
Cofactor of element 3 = (-1)¹⁺³(32) = 32

Step 5: Continue the procedure for all the elements to find the cofactor matrix of A,
i.e., Cofactor Matrix of A =  [Tex]\begin{bmatrix} -4&-33&32\\ 6&-9&4\\-2&16&-10 \end{bmatrix}[/Tex]

Transpose of a Matrix

Transpose of a matrix is the matrix that is formed by changing the rows and columns of the matrix with each other. The transpose of the matrix A is denoted as A^T or A^‘. If the order of the matrix A is m×n, then the order of the transpose matrix is n×m.

Learn more about, Transpose of a Matrix

How to find the Adjoint of a Matrix?

To find the Adjoint of a Matrix, first, we have to find the Cofactor of each element, and then take 2 more steps. See below the steps,

Step 1: Find the Cofactor of each element present in the matrix.
Step 2: Create another matrix with the cofactors as its elements.
Step 3: Now find the transpose of the matrix which comes from after Step 2.

How to find the Adjoint of a 2×2 Matrix

Let’s consider an example for understanding the method to find the adjoint of a 2×2 Matrix.

Example: Find the Adjoint of[Tex]\bold{\text{A} =\begin{bmatrix}2&3\\ 4&5 \end{bmatrix}}        [/Tex].

Solution:

Given matrix is [Tex]\text{A} =\begin{bmatrix}2&3\\ 4&5 \end{bmatrix}[/Tex]

Step 1: Find the Cofactor of each element.
Cofactor of element at A[1, 1]: 5
Cofactor of element at A[1, 2]: -4
Cofactor of element at A[2, 1]: -3
Cofactor of element at A[2, 2]: 2

Step 2: Create matrix from Cofactors
i.e.,[Tex]\bold{\begin{bmatrix}5&-4\\ -3&2 \end{bmatrix}}[/Tex]

Step 3: Transpose of Cofactor matrix,
[Tex]\bold{Adj(A) =  \begin{bmatrix}5&-3\\ -4&2 \end{bmatrix}}[/Tex]

How to find the Adjoint of a 3×3 Matrix

Let’s take an example of a 3×3 Matrix to understand how to calculate the Adjoint of that matrix.

Example: Find the Adjoint of[Tex]\bold{A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}}        [/Tex].

Solution:

[Tex]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}[/Tex]

Step 1: Find the Cofactor of each element.

C₁₁ = (-1)¹⁺¹[Tex]\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix}[/Tex] = (45 – 48) = -3
C₁₂ = (-1)¹⁺²[Tex]\begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix}[/Tex] = -(36 – 42) = 6
C₁₃ = (-1)¹⁺³[Tex]\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}[/Tex] = (32 – 35) = -3
C₂₁ = (-1)²⁺¹[Tex]\begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix}[/Tex] = -(18 – 24) = 6
C₂₂= (-1)²⁺²[Tex]\begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix}[/Tex] = (9 – 21) = -12
C₂₃ = (-1)²⁺³[Tex]\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}[/Tex] = -(8 – 14) = 6
C₃₁= (-1)³⁺¹[Tex]\begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix}[/Tex] = (12 – 15) = -3
C₃₂ = (-1)³⁺²[Tex]\begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix}[/Tex] = -(6 – 12) = 6
C₃₃ = (-1)³⁺³[Tex]\begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix}[/Tex] = (5 – 8) = -3

Step 2: Create matrix from Cofactors
[Tex]C = \begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \\ \end{bmatrix}[/Tex]

Step 3: Transpose of Matrix C to adjoint of given matrix, rows become columns and columns become row.
[Tex]\operatorname{adj}(A) = C^{T}= \begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \\ \end{bmatrix}[/Tex]

Which is adjoint of given matrix A.

Properties of the Adjoint of a Matrix

Adjoints of a matrix have various properties, some of which are as follows:

• A(Adj A) = (Adj A)A = |A| I_n
• Adj(BA) = (Adj B) (Adj A)
• |Adj A| = |A|^n-1
• Adj(kA) = k^n-1(Adj A)

Finding the Inverse Using the Adjoint of a Matrix

Finding the inverse is one of the important applications of the Adjoint of the Matrix. To find the inverse of a Matrix using the Adjoint, we can use the following steps:

Step 1: Find the determinant of the matrix.
Step 2: If the determinant is zero, then the matrix is not invertible, and there is no inverse.
Step 3: If the determinant is non-zero, then find the adjoint of the matrix.
Step 4: Divide the adjoint of the matrix by the determinant of a matrix.
Step 5: The result of Step 4 is the Inverse of the given Matrix.

Example: Find the inverse of[Tex]\bold{A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}}        [/Tex].

Solution:

Given matrix [Tex]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}[/Tex]

|A| = 1(45 – 48) -2(36 – 42) + 3(32 – 35)
⇒ |A| = -3 -2(-6) + 3(-3)
⇒ |A| = -3 + 12 – 9 = 0

Thus, inverse of A doesn’t exist.

Learn more about a Matrix

Solved Examples of the Adjoint of a Matrix

Example 1: Find the Adjoint of the given matrix[Tex]A =\begin{bmatrix} 1 & 2 & 3\\ 7 & 4 & 5 \\ 6 & 8 & 9 \end{bmatrix}           [/Tex].

Solution:

Step 1: To find the cofactor of each element
To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of elements at A[1, 1] = 1: [Tex]+\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix}         [/Tex] = +(4 x 9 – 8 x 5) = -4
Cofactor of elements at A[1, 2] = 2: [Tex]-\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix}         [/Tex] = -(7 x 9 – 6 x 5) = -33
Cofactor of elements at A[1, 3] = 3: [Tex]+\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}         [/Tex] = +(7 x 8 – 6 x 4) = 32
Cofactor of elements at A[2, 1] = 7: [Tex]-\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix}         [/Tex] = -(2 x 9 – 8 x 3) = 6
Cofactor of elements at A[2, 2] = 4: [Tex]+\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix}         [/Tex] = +(1 x 9 – 6 x 3) = -9
Cofactor of elements at A[2, 3] = 5: [Tex]-\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix}         [/Tex] = -(1 x 8 – 6 x 2) = 4
Cofactor of elements at A[3, 1] = 6: [Tex]+\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}         [/Tex] = +(2 x 5 – 4 x 3) = -2
Cofactor of elements at A[3, 2] = 8: [Tex]-\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix}         [/Tex] = -(1 x 5 – 7 x 3) = 16
Cofactor of elements at A[3, 3] = 9: [Tex]+\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix}         [/Tex] = +(1 x 4 – 7 x 2) = -10

The matrix looks like with the cofactors:

[Tex]A =\begin{bmatrix} +\begin{bmatrix} 4 & 5 \\ 8 & 9 \end{bmatrix} & -\begin{bmatrix} 7 & 5 \\ 6 & 9 \end{bmatrix} & +\begin{bmatrix} 7 & 4 \\ 6 & 8 \end{bmatrix}\\ \\ -\begin{bmatrix} 2 & 3 \\ 8 & 9 \end{bmatrix} & +\begin{bmatrix} 1 & 3 \\ 6 & 9 \end{bmatrix} & -\begin{bmatrix} 1 & 2 \\ 6 & 8 \end{bmatrix} \\ \\         +\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} & -\begin{bmatrix} 1 & 3 \\ 7 & 5 \end{bmatrix} & +\begin{bmatrix} 1 & 2 \\ 7 & 4 \end{bmatrix} \end{bmatrix}[/Tex]

The final cofactor matrix:
[Tex]A =\begin{bmatrix} -4 & -33 & 32\\ 6 & -9 & 4 \\ -2 & 16 & -10 \end{bmatrix}[/Tex]

Step 2: Find the transpose of the matrix obtained in step 1
[Tex]adj(A) =\begin{bmatrix} -4 & 6 & -2\\ -33 & -9 & 16 \\ 32 & 4 & -10 \end{bmatrix}[/Tex]

This is the Adjoint of the matrix.

Example 2: Find the Adjoint of the given matrix[Tex]A =\begin{bmatrix} -1 & -2 & -2\\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}           [/Tex].

Solution:

Step 1: To find the cofactor of each element

To find the cofactor of each element, we have to delete the row and column of each element one by one and take the present elements after deleting.

Cofactor of element at A[1, 1] = -1: [Tex]+\begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}         [/Tex] = +(1 x 1 – (-2) x (-2)) = -3
Cofactor of elements at A[1, 2] = -2: [Tex]-\begin{bmatrix} 2 & -2 \\ 2 & 1 \end{bmatrix}         [/Tex] = -(2 x 1 – 2 x (-2)) = -6
Cofactor of elements at A[1, 3] = -2: [Tex]+\begin{bmatrix} 2 & 1 \\ 2 & -2 \end{bmatrix}         [/Tex] = +(2 x (-2) – 2 x 1) = -6
Cofactor of elements at A[2, 1] = 2: [Tex]-\begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}         [/Tex] = -((-2) x 1 – (-2) x (-2)) = 6
Cofactor of elements at A[2, 2] = 1: [Tex] +\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}         [/Tex]  = +((-1) x 1 – 2 x (-2)) = 3
Cofactor of elements at A[2, 3] = -2: [Tex]-\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}         [/Tex]  = -((-1) x (-2) – 2 x (-2)) = -6
Cofactor of elements at A[3, 1] = 2: [Tex]+\begin{bmatrix} -2 & -2 \\ 1 & -2 \end{bmatrix}         [/Tex] = +((-2) x (-2) – 1 x (-2)) = 6
Cofactor of elements at A[3, 2] = -2: [Tex]-\begin{bmatrix} -1 & -2 \\ 2 & -2 \end{bmatrix}         [/Tex]  = -((-1) x (-2) – 2 x (-2)) = -6
Cofactor of elements at A[3, 3] = 1: [Tex]+\begin{bmatrix} -1 & -2 \\ 2 & 1 \end{bmatrix}         [/Tex] = +((-1) x (-1)- 2 x (-2)) = 3

The final cofactor matrix:
[Tex]A =\begin{bmatrix} -3 & -6 & -6\\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}[/Tex]

Step 2: Find the transpose of the matrix obtained in Step 1
[Tex]adj(A) =\begin{bmatrix} -3 & 6 & 6\\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}[/Tex]

This is the Adjoint of the matrix.