Inverse of the following matrix. $$M=\left[\begin{array}{cc}
       (m+2) &  2\cdot1{^\prime}_m \\\
       2\cdot1_m  &  I_{m}+J_{m,m}\\
        \end{array}\right]
$$
Is there a closed form of  $M^{-1}$?
Where $I_m$ is an identity matrix, $1_m$ is a column vector of ones and $J_{m,m}$ is a matrix of ones. 
 A: Since $\left( {{\bf I}_{\,m}  + {\bf J}_{\,m} } \right)$ has a quite compact inverse, which is easy to demonstrate to be:
$$
\left( {{\bf I}_{\,m}  + {\bf J}_{\,m} } \right)^{\,{\bf  - 1}}  = \left( {{\bf I}_{\,m}  - {1 \over {m + 1}}{\bf J}_{\,m} } \right)
$$
then we can proceed in two alternative ways:


*

*use Blockwise Inversion, which is
$$
\left[ {\matrix{
   {\bf A} & {\bf B}  \cr 
   {\bf C} & {\bf D}  \cr 
 } } \right]^{\,\,{\bf  - 1}}  = \left[ {\matrix{
   {\left( {{\bf A - BD}^{\,{\bf  - 1}} {\bf C}} \right)^{\,{\bf  - 1}} } & {{\bf  - }\left( {{\bf A - BD}^{\,{\bf  - 1}} {\bf C}} \right)^{\,{\bf  - 1}} {\bf BD}^{\,{\bf  - 1}} }  \cr 
   {{\bf  - D}^{\,{\bf  - 1}} {\bf C}\left( {{\bf A - BD}^{\,{\bf  - 1}} {\bf C}} \right)^{\,{\bf  - 1}} } & {{\bf D}^{\,{\bf  - 1}} {\bf  + D}^{\,{\bf  - 1}} {\bf C}\left( {{\bf A - BD}^{\,{\bf  - 1}} {\bf C}} \right)^{\,{\bf  - 1}} {\bf BD}^{\,{\bf  - 1}} }  \cr 
 } } \right]
$$

*standing the formula for the inverse of $\left( {{\bf I}_{\,m}  + {\bf J}_{\,m} } \right)$ and the fact
that $\mathbf M$ is symmetric, we can assume that 
$$
{\bf M}_{\,m} ^{\,\,{\bf  - 1}}  = \left[ {\matrix{
   a & {b\overline {\bf 1} _{\,m} }  \cr 
   {b{\bf 1}_{\,m} } & {c{\bf I}_{\,m}  + d{\bf J}_{\,m} }  \cr 
 } } \right]
$$
then, imposing that
$$
\eqalign{
  & {\bf I}_{\,m}  = {\bf M}_{\,m} {\bf M}_{\,m} ^{\,\,{\bf  - 1}}  =   \cr 
  &  = \left[ {\matrix{
   {m + 2} & {2\overline {\bf 1} _{\,m} }  \cr 
   {2{\bf 1}_{\,m} } & {{\bf I}_{\,m}  + {\bf J}_{\,m} }  \cr 
 } } \right]\left[ {\matrix{
   a & {b\overline {\bf 1} _{\,m} }  \cr 
   {b{\bf 1}_{\,m} } & {c{\bf I}_{\,m}  + d{\bf J}_{\,m} }  \cr 
 } } \right] =   \cr 
  &  = \left[ {\matrix{
   {\left( {m + 2} \right)a + 2mb} & {\left( {\left( {m + 2} \right)b + 2c + 2md} \right)\overline {\bf 1} _{\,m} }  \cr 
   {\left( {2a + \left( {m + 1} \right)b} \right){\bf 1}_{\,m} } & {c\,{\bf I}_{\,m}  + \left( {2b + \left( {m + 1} \right)d + c} \right){\bf J}_{\,m} }  \cr 
 } } \right] \cr} 
$$
we arrive at:
$$
\left\{ \matrix{
  \left( {m + 2} \right)a + 2mb = 1 \hfill \cr 
  \left( {m + 2} \right)b + 2c + 2md = 0 \hfill \cr 
  2a + \left( {m + 1} \right)b = 0 \hfill \cr 
  c = 1 \hfill \cr 
  2b + c + \left( {m + 1} \right)d = 0 \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  a = \left( {m + 1} \right)/\left( {m^{\,2}  - m + 2} \right) \hfill \cr 
  b =  - 2/\left( {m^{\,2}  - m + 2} \right) \hfill \cr 
  c = 1 \hfill \cr 
  d = \left( {2 - m} \right)/\left( {m^{\,2}  - m + 2} \right) \hfill \cr}  \right.
$$

