I put my derivations in this image here: [![enter image description here][1]][1]
I am just using gaussian elimination by integrating all row elements. Is this acceptable?
With this approach, we only need to invert the hessian once for any nxn matrix. Then we can plug in our values.
No, integration is not a row operation in the elimination.
Added some details: To compute the inverse, the procedures is to multiply some matrices of row operations on both sides, if H is invertible, then the left block can be reduced to the identity matrix I.
$$H | I$$ $$E_n E_{n-1} \cdots E_1 H | E_n E_{n-1} \cdots E_1I$$ $$I|E_n E_{n-1} \cdots E_1$$
If the left block: $E_n E_{n-1} \cdots E_1 H = I$. Then, $H^{-1} = E_n E_{n-1} \cdots E_1 $.
You may go back to how you solve system of equations originally, Try solving this, g should be left parts of the inverse,
$ \begin{bmatrix} f_{11} & f_{12}\\ f_{21} & f_{22} \end{bmatrix} \begin{bmatrix} g_1\\ g_2 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix} $
$ \begin{bmatrix} g_1 f_{11} + g_2 f_{12}\\ g_1 f_{21} + g_2 f_{22} \end{bmatrix}= \begin{bmatrix} 1\\ 0 \end{bmatrix} $
Then try to integrate the first row, note that $g_1$ and $g_2$ is not just a constant, it is a function of (x,y), that's why you cant integrate both rows like that.
Moreover, when you integrate on both rows, the constants c are arbitrary, the values of two rows are not the same, and you still need to eliminate the constatnt c to get the identity matrix.