How to find the determinant of the following block matrix?
$$\begin{pmatrix} \lambda I-A & -A &-J\\ -A& \lambda I & 0\\ -J& 0 & \lambda I\end{pmatrix}$$
where $A$ is a $n \times n$ matrix and $\lambda$ is a scalar. $J$ is a square matrix of order $n$.
I tried using row and column operations like using $R_1\to R_1+R_2$ but I am not getting anything fruitful. What is the way out? Is there any special trick to do this?
Using the Schur complement formula for the determinant of a block matrix,$$ \det \begin{pmatrix} X & Y\\ Z& W\end{pmatrix} = \det W \det(X- Y W^{-1} Z),$$ with the lower right $2n\times 2n$ block to be $W$ as it's simple to invert, I find that $$\det \begin{pmatrix} \lambda I-A & -A &-J\\ -A& \lambda I & 0\\ -J& 0 & \lambda I\end{pmatrix} = \lambda^n \det (\lambda^2 I - \lambda A - A^2 - J^2).$$
Edit; I applied the determinant formula differently and found $$\det \begin{pmatrix} \lambda I-A & -A &-J\\ -A& \lambda I & 0\\ -J& 0 & \lambda I\end{pmatrix} = \frac{\det(J^2)\det (2\lambda^2 I - 2A^2 - \lambda A)}{\det(\lambda I -A)},$$ which you might prefer depending on your application, @Math_Freak