How to find determinant of the following $3\times 3$ matrix? [closed]

Solution 1:

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