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. 

Note that
$$A = (a-b)I + b e e^T$$
where $I$ is the identity matrix, $e = \begin{bmatrix} 1&1&1& \cdots &1 \end{bmatrix}^T$. We have
$$B = A+ \dfrac{ee^T}n = (a-b)I + \left(b + \dfrac1n\right)ee^T$$
Now make use of Sherman-Morrison Woodbury formula to compute $AB^{-1}A$ and conclude what you want.
I have added the details below. Hoever you mouse over the gray area below for more details.

Using Sherman-Morrison Woodbury formula, we get that$$B^{-1} = \dfrac{I}{a-b} - \dfrac{ee^T}{(a-b)^2 + n(a-b)}$$ Hence, $$AB^{-1} = I + \dfrac{b}{a-b} ee^T - \dfrac{(a-b)ee^T}{(a-b)^2+n(a-b)} - \dfrac{bn ee^T}{(a-b)^2+n(a-b)}\\ = I + \left(\dfrac{b}{a-b} - \dfrac{a}{(a-b)^2+n(a-b)}\right)ee^T\\ = I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T$$$$AB^{-1}A = \left(I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T \right) \left( (a-b)I + b e e^T\right)\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}(a-b) + b + \color{blue}{\underbrace{\dfrac{nb(ab-b^2+nb-a)}{(a-b)^2+n(a-b)}}_{\text{This term makes use of the fact that $e^Te = n$}}}\right) ee^T\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)} \color{red}{\underbrace{(a-b + nb)}_{\text{This term is zero}}} + b\right) ee^T\\= (a-b)I + bee^T = A$$

