Processing math: 100%

$\LARGE adj(M)$ = Adjoint Mtrix of M = Transposed Cofactor Matrix of M = $\LARGE Cofac(M)^t$

Got it !

Adjoint matrix of a 3 x 3 matrix

$\LARGE adj \begin{bmatrix} \color{black}1 & \color{black}2 & \color{black}-1 \\\color{black}2 &\color{black}1&\color{black}2\\\color{black}-1 &\color{black}2&\color{black}1 \end{bmatrix} =$	$\LARGE \begin{bmatrix} \color{black}-3 & \color{black}-4 & \color{black}5 \\\color{black}-4 &\color{black}0&\color{black}-4\\\color{black}5 &\color{black}-4&\color{black}-3 \end{bmatrix}^t =$ $\LARGE \begin{bmatrix} \color{black}-3 & \color{black}-4 & \color{black}5 \\\color{black}-4 &\color{black}0&\color{black}-4\\\color{black}5 &\color{black}-4&\color{black}-3 \end{bmatrix}$
$\LARGE adj \begin{bmatrix} \color{black}1 & \color{black}2 & \color{black}3 \\\color{black}0 &\color{black}4&\color{black}5\\\color{black}0 &\color{black}0&\color{black}6 \end{bmatrix} =$	$\LARGE \begin{bmatrix} \color{black}24 & \color{black}0 & \color{black}0 \\\color{black}-12 &\color{black}6&\color{black}0\\\color{black}-2 &\color{black}-5&\color{black}4 \end{bmatrix}^t =$ $\LARGE \begin{bmatrix} \color{black}24 & \color{black}-12 & \color{black}-2 \\\color{black}0 &\color{black}6&\color{black}-5\\\color{black}0 &\color{black}0&\color{black}4 \end{bmatrix}$
$\LARGE adj \begin{bmatrix} \color{black}1 & \color{black}0 & \color{black}0 \\\color{black}0 &\color{black}1&\color{black}0\\\color{black}0 &\color{black}0&\color{black}1 \end{bmatrix} =$	$\LARGE \begin{bmatrix} \color{black}1 & \color{black}0 & \color{black}0 \\\color{black}0 &\color{black}1&\color{black}0\\\color{black}0 &\color{black}0&\color{black}1 \end{bmatrix}^t =$ $\LARGE \begin{bmatrix} \color{black}1 & \color{black}0 & \color{black}0 \\\color{black}0 &\color{black}1&\color{black}0\\\color{black}0 &\color{black}0&\color{black}1 \end{bmatrix}$