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.

## Finding Adjoint of a Matrix

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

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 3x3 Matrix

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

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

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

### Inverse Matrix

 Matrix Augment Matrix Column Matrix Diagonal of Matrix How to do Matrix Multiplication Identity of a Matrix Inverse of a Matrix Matrix Addition Matrix Equations Properties of Matrix The Determinant of a Matrix Transpose a Matrix
 Calculate Matrix Matrix Multiply Adding Matrix Calculator