The cofactor

of a square matrix
$A$
is the determinant of the submatrix obtained by deleting the ith row and jth column, multiplied by
$(-1)^{i+j}$
.
For example, if
$A=\left( \begin{array}{ccc} 2 & 3 & 0 \\ 0 & 1 & 5 \\ 4 & -2 & 1 \end{array} \right)$

then
$A_{11}=(-1)^2 \left| \begin{array}{cc} 1 & 5 \\ -2 & 1 \end{array} \right| = 1 \times 1 - 5 \times (-2)=11$

$A_{12}=(-1)^3 \left| \begin{array}{cc} 0 & 5 \\ 4 & 1 \end{array} \right| = -(0 \times 1 - 5 \times 4)=19$

$A_{13}=(-1)^4 \left| \begin{array}{cc} 0 & 1 \\ 4 & -2 \end{array} \right| = 0 \times -2 - 1 \times 4=-4$

$A_{21}=(-1)^3 \left| \begin{array}{cc} 3 & 0 \\ -2 & 1 \end{array} \right| = -(3 \times 1 - 0 \times -2)=-3$

$A_{22}=(-1)^4 \left| \begin{array}{cc} 2 & 0 \\ 4 & 1 \end{array} \right| = 2 \times 1 - 0 \times 4=2$

$A_{23}=(-1)^5 \left| \begin{array}{cc} 2 & 3 \\ 4 & -2 \end{array} \right| = -( 2 \times -2 - 3 \times 4)=16$

$A_{31}=(-1)^4 \left| \begin{array}{cc} 3 & 0 \\ 1 & 5 \end{array} \right| = 3 \times 5 - 0 \times 1=15$

$A_{32}=(-1)^5 \left| \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \right| = -(2 \times 5 - 0 \times 0)=-10$

$A_{33}=(-1)^6 \left| \begin{array}{cc} 2 & 3 \\ 0 & 1 \end{array} \right| = 2 \times 1 - 3 \times 0=2$

The matrix of cofactors is
$\left( \begin{array}{ccc} 11 & 19 & -4 \\ -3 & 2 & 16 \\ 15 & -10 & 2 \end{array} \right)$

The adjoint matrix is the transpose of the matrix of cofactors, and is equal to
$\left( \begin{array}{ccc} 11 & -3 & 15 \\ 19 & 2 & -10 \\ 4 & 16 & 2 \end{array} \right)$