considering a $2\times 2$ matrix $\bf S$,
\begin{equation} {\bf S} = \begin{bmatrix} \frac{\partial}{\partial{t}} & \kappa\nabla .\\ 1/\rho \nabla . & \frac{\partial}{\partial{t}} \end{bmatrix} \end{equation}
I want to get the adjoint of $\bf S$, $$\bf S^{*}$$ for that I am using cofactor matrix and then I transpose this matrix (METHOD 1 to get adjoint matrix).
\begin{equation} {\bf S}_{\text{cof}} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \end{equation} as $$a_{ij}=(-1)^{i+j}\det\bf S_{ij}$$
\begin{eqnarray} a_{11}&=&\frac{\partial}{\partial{t}}\\ a_{12}&=&-1/\rho \nabla .\\ a_{21}&=&-\kappa\nabla . \\ a_{22}&=&\frac{\partial}{\partial{t}} \end{eqnarray}
In matrix notation we have,
\begin{equation} {\bf S}_{\text{cof}} = \begin{bmatrix} \frac{\partial}{\partial{t}} & -1/\rho \nabla \\ -\kappa\nabla . & \frac{\partial}{\partial{t}} \end{bmatrix} \end{equation}
Transposing the matrix $$\bf S_{\text{cof}}$$ we get:
\begin{equation} {\bf S}^{*}= \begin{bmatrix} \frac{\partial}{\partial{t}} & -\kappa\nabla . \\ -1/\rho \nabla .& \frac{\partial}{\partial{t}} \end{bmatrix} \end{equation}
This is the way I am doing but I saw some other papers doing something different, the procedure other people are doing is transposing $\bf S$ and change the signal of every first derivative (METHOD 2 to get adjoint matrix), in this sense the adjoint of $\bf S$ is:
\begin{equation} {\bf S}^{*} = \begin{bmatrix} - \frac{\partial}{\partial{t}} & -1/\rho \nabla \\ -\kappa\nabla . & - \frac{\partial}{\partial{t}} \end{bmatrix} \end{equation}
Which method is right? Am I doing something wrong? If you have any documentation please share it.
Thank you, Rafael