Let $r < n^2$, denote by $I_n$ the $n \times n$ identity matrix and consider the block matrix
$$A = \left[ \begin{array}{cccccc} 0 & I_n & I_n & \ldots & I_n & I_n\\ I_n & 0 & I_n & \ldots & I_n & I_n\\ I_n & I_n & 0 & \ldots & I_n & I_n\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ I_n & I_n & I_n & \ldots & I_n & 0\\ \end{array} \right]$$ composed by $r \times r$ blocks. The zeros in the matrix represents a $n \times n$ block matrix filled with zeros.
Note the pattern, each block matrix in the diagonal of $A$ is a $n \times n$ zero matrix and the other blocks are identities. I'm interested in computing or at least bounding (by above) the norm
$$\left\| \left[ \begin{array}{ccc} 0 & A & A\\ A & 0 & A\\ A & A & 0\\ \end{array}\right]^{-1} \right\|$$ where each zero now is a $rn \times rn$ block matrix filled with zeros.
For some values for $r,n$ I tested, this matrix was invertible. I suspect it always is the case, but can't prove. I need a help to prove this and to bound this norm, which can be the usual spectral norm or the Frobenius norm.
Thank you very much.
PS: The condition $r < n^2$ comes from the original problem which lead me to this matrix. You can ignore it if not helpful.
Let $\mathbf1_k=(1,1,\ldots,1)^T\in\mathbb R^k$ denotes the $k$-vector of ones. Your block matrix is just $(\mathbf1_3\mathbf1_3^T-I_3)\otimes (\mathbf1_r\mathbf1_r^T-I_r)\otimes I_n$. Hence its inverse is $(\frac12\mathbf1_3\mathbf1_3^T-I_3)\otimes (\frac1{r-1}\mathbf1_r\mathbf1_r^T-I_r)\otimes I_n$ when $r>1$ and the spectral norm of this inverse is $1$.