Determinant of block matrices I am having the following block matrix
$$\left[\begin{array}{cccc}
         \mathbf{A+B} & \mathbf{B} & \cdots & \mathbf{B} \\
         \mathbf{B} & \mathbf{A+B} & \cdots & \mathbf{B} \\
         \vdots     &  \vdots      & \ddots & \vdots\\
         \mathbf{B} & \mathbf{B} & \cdots & \mathbf{A+B}
       \end{array}\right]$$
where $\mathbf{A}$ and $\mathbf{B}$ are full rank, symmetric square matrices. There are $n$ blocks in each direction. I want to obtain the determinant of the block matrix.
I play with some examples and the determinants seems to be 
$$\det(\mathbf{A})^{n-1}\det(\mathbf{A}+n\mathbf{B})$$
May I ask whether this is correct or not, and is there any proof?
 A: the above-mentioned link uses to end up the proof a topological argument of density. Here is a simple purely algebraic proof, based on operations on rows and on columns:
Substracting the last block-row from the $n-1$ first block-rows yields:
$$\begin{bmatrix}
A&0&0&\dots&0&-A\\
0&A&0&\dots&0&-A\\
0&0&A&\dots&0&-A\\
\vdots&&&&&\vdots\\
0& 0&0&\dots&A&-A \\
B&B&B&\dots&B&A+B
\end{bmatrix}$$
Now add each of the $n-1$ first columns to the last one, to get:
$$\begin{bmatrix}
A&0&0&\dots&0&0\\
0&A&0&\dots&0&0\\
0&0&A&\dots&0&0\\
\vdots&&&&&\vdots\\
0& 0&0&\dots&A&0\\
B&B&B&\dots&B&A+nB
\end{bmatrix}$$
We have a lower block-triangular matrix. Its determinant is the product of the determinants of the diagonal blocks:
$$\lvert A\rvert^{n-1}\cdot\lvert A+nB\rvert.$$
A: Your matrix is equal to $E\otimes B+I\otimes A$ (where $E$ is the all-one matrix), which is similar to $nE_{11}\otimes B+I\otimes A=\operatorname{diag}(nB+A,\,A,\,\ldots,\,A)$ (where $E_{11}$ is the matrix whose only nonzero entry is a $1$ at the $(1,1)$-th position). Hence the result.
