Adjoint Matrix can be found by forming the cofactor matrix and then finding its transpose. Adjoint Matrix is defined as follows:

If $A_{n\times n}$= $\left [ a_{ij} \right ]_{n\times n}$, then adj A = $\left [ c_{ij} \right ]_{n\times n}^T$, where Cij = (-1)i+j Mij and Mij indicates minors of the elements.

In order to find the adjoint of a matrix A, which is a square matrix of order nxn, we have to follow procedure given below:

  • Calculate cofactors of each element of the matrix using minors.
  • Form cofactor matrix from the calculated cofactors.
  • Find transpose matrix of the cofactor matrix.

Adjoint of a 2x2 matrix can be found by following the steps given below:

B = $\begin{bmatrix}
2 &4 \\
-6 &8
\end{bmatrix}$

First, we have to calculate the cofactors of each element. So,
c11=(-1)1+18 = 8 
c12=(-1)1+2 (-6)= 6 
c21=(-1)2+1 (4) = -4
c22=(-1)2+2 (2) = 2

Now, form a cofactor matrix of B.
C = $\begin{bmatrix}
8 &6 \\
-4 &2
\end{bmatrix}$ 

adj B = $C^T$ = $\begin{bmatrix}
8 &6 \\
-4 &2
\end{bmatrix}^T$

adj B = $\begin{bmatrix}
8 &-4 \\
6 &2
\end{bmatrix}$   

Adjoint of a 2x2 matrix can be found by following the steps given below:

A = $\begin{bmatrix}
6 &2  &1 \\
1 &3  &5 \\
4 &4  &1
\end{bmatrix}$

For adj A,  we follow some steps:
Calculate cofactors of each element
Cofactor of 6 (element of 1st row and 1st column) = (-1)1+1 $\begin{vmatrix}
3 &5 \\
4 &1
\end{vmatrix}$


                                                                                   = 1(3 - 20) = -17

Cofactor of 2 (element of 1st row and 2nd column) = (-1)1+2 $\begin{vmatrix}
1 &5 \\
4 &1
\end{vmatrix}$

                                                                                  = (-1)(1- 20) =19

Cofactor of 1 (element of 1st row and 3rd column) = (-1)4 $\begin{vmatrix}
1 &3 \\
4 &41
\end{vmatrix}$

                                                                                    = 1(4 -12) = -8

cofactor of 1 (element of 2nd row and 1st column) = (-1)3 $\begin{vmatrix}
2 &1 \\
4 &1
\end{vmatrix}$

                                                                                     = (-1) ( 2-4) = 2

Cofactor of 3 (element of 2nd row and 2ndcolumn) = (-1)4$\begin{vmatrix}
6 &1\\
4 &1
\end{vmatrix}$  
            
                                                                                    = 1(6 - 4) = 2

Cofactor of 5 (element of 2nd row and 3rdcolumn) = (1-)5 $\begin{vmatrix}
6 &2 \\
4 &4
\end{vmatrix}$

                                                                                     = (-1) ( 24 - 8) = 16

Cofactor of 4 (element of 3rd row and 1st column)  = (-1)4 $\begin{vmatrix}
2 &1 \\
3 &5
\end{vmatrix}$

                                                                                    = 1(10 - 3) = 7
   
Cofactor of 4 (element of 3rd row and 2nd column) = (-1)5 $\begin{vmatrix}
6 &1 \\
1 &5
\end{vmatrix}$

                                                                                      = (-1) (30 - 1) = -29
         
Cofactor of 4 (element of 3rd row and 3rdcolumn)  = (-1)6 $\begin{vmatrix}
6 &2 \\
1 &3
\end{vmatrix}$

                                                                                    = 1( 18 - 2) = 16

Now, form a cofactor matrix C using these values.
C = $\begin{bmatrix}
-17 &19  &-8 \\
2 &2  &16 \\
7 &-29 &16
\end{bmatrix}$

Then, adj B = $C^T$ = $\begin{bmatrix}
-17 &2  &7 \\
19 &2  &-29 \\
-8 &16 &16
\end{bmatrix}$