Determinant of $N \times N$ matrix

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.

$$\det(\boldsymbol J (E_i) - \lambda \mathbb I ) = \begin{pmatrix} A_1 -\lambda & \dots & \phi_{1i} & \dots & 0 \\ \vdots & \ddots & \vdots & & \vdots \\ -c_{i1} & \dots & \color{red}{A_i -\lambda} & \dots & -c_{iN}\\ \vdots & & \vdots & \ddots & \vdots \\ 0 & \dots & \phi_{Ni} & \dots & A_N -\lambda \end{pmatrix}$$ 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 $i=3$, $$\det(\boldsymbol J (E_i) - \lambda \mathbb I ) = \begin{pmatrix} D_1 & 0 & V_1 & 0 \\ 0 & D_2 & V_2&0 \\ H_1 & H_2 & D_3 & H_4\\ 0 & 0 & V_4 & D_4 \\ \end{pmatrix}$$

• If from diagonal entries vertical /horizontal elements in that row/column are nonzero, that seems to go against the two $0$ entries upper right and lower left. Apr 7, 2017 at 11:28
• @coffeemath Apologies for my poor wording, please see the update! Apr 7, 2017 at 11:30
• So, wait, your original matrix is like $[[1, 0, 2, 0, 0], [0, 3, 0, 0, 0], [4, 0, 5, 0, 6], [0, 0, 0, 7, 0], [0, 0, 8, 0, 9]]$? Apr 7, 2017 at 11:38
• Could the upper right and lower left entries, now labeled as $0,$ also be nonzero in the case $i \neq 1,N$? Apr 7, 2017 at 11:42
• Step 1: Change basis from $e_1, e_2, \ldots e_n$ to $e_i, e_2, \ldots, e_1, \ldots e_n$, exchanging rows $1$ and $i$ and columns $1$ and $i$. Then the first row and column become nonzero, and the remainder of the matrix is diagonal. That'll at least simplify the notation without changing the determinant. Apr 7, 2017 at 11:44

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. $$\left(A_i-\lambda + \sum_{k\ne i}\frac{c_k\phi_{ki}}{A_k-\lambda}\right) \prod_{k\ne i}(A_k-\lambda).$$