<s>[INST] How to show $AB^{-1}A=A$ 
Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$
  where $a \neq b$ and $a + (n - 1)b = 0$.
  Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n \times 1$ vector. Show that $AB^{-1}A=A$

Trial: Here $B^{-1}=\frac{1}{\alpha-\beta}I_n-\frac{\beta~11'}{(\alpha-\beta)(\alpha+(n-1)\beta)}$ Where $(\alpha-\beta)=(a+\frac{1}{n}-b-\frac{1}{n})=(a-b)$ and $\alpha+(n-1)\beta=1$ So, $B^{-1}=\frac{1}{a-b}[I_n-(b+\frac{1}{n})11']$. Then pre and post multipling $A$ I can't reach to the desire result . Please help. 
 [/INST] Here is another answer of mine. Let $E={\mathbf 1}{\mathbf 1}^T$. By Sherman-Morrison formula,
\begin{align*}
&B = (a-b)I + \left(b + \frac1n\right){\mathbf 1}{\mathbf 1}^T\\
\Rightarrow&B^{-1} = \frac{I}{a-b} + cE
\end{align*}
for some constant $c$. We will see that the exact value of $c$ is unimportant. Now, note that
\begin{align*}
&A = (a-b)I + bE\\
\Rightarrow&AE = (a-b)E + bE^2 = (a-b)E + nbE = 0.
\end{align*}
Therefore,
\begin{align*}
AB^{-1}A
&= A\left(\frac{I}{a-b} + cE\right)[(a-b)I + bE]\\
&= \left(\frac{A}{a-b} + cAE\right)[(a-b)I + bE]\\
&= \frac{A}{a-b}[(a-b)I + bE]\\
&= A.
\end{align*}
</s>