I have the following matrix which I need to find the determinant of. I am not too sure of how to proceed. Here is my working so far.
\begin{equation}
\det(\boldsymbol J (E_i) - \lambda \mathbb I ) =
\begin{pmatrix}
A_1 -\lambda & \dots & \phi_{1i} & \dots & 0 \\
\vdots & \ddots & \vdots & & \vdots \\
-c_{i1} & \dots & \color{red}{A_i -\lambda} & \dots & -c_{iN}\\
\vdots & & \vdots & \ddots & \vdots \\
0 & \dots & \phi_{Ni} & \dots & A_N -\lambda
\end{pmatrix}
\end{equation}
The matrix has a diagonal given by $A_j -\lambda$. From the central red element there are vertically and horizontally non-zero elements. All other elements are zero exactly.
I am really not sure how to find the determinant from here and any help or pointers would be greatly appreciated!
Edit
For instance if $N$ where to equal 4 we might have the following case if $i=3$,
\begin{equation}
\det(\boldsymbol J (E_i) - \lambda \mathbb I ) =
\begin{pmatrix}
D_1 & 0 & V_1 & 0 \\
0 & D_2 & V_2&0 \\
H_1 & H_2 & D_3 & H_4\\
0 & 0 & V_4 & D_4 \\
\end{pmatrix}
\end{equation}
Answer
Edit. Let $B$ the submatrix obtained by deleting the $i$-th row and $i$-th column of the given matrix. Then the required determinant is the product of determinant of the Schur complement of $B$ and $\det B$, i.e.
$$
\left(A_i-\lambda + \sum_{k\ne i}\frac{c_k\phi_{ki}}{A_k-\lambda}\right) \prod_{k\ne i}(A_k-\lambda).
$$
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