<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] Writing replacing $a$ with $(1-n)b$ and $\mathbf{1}\mathbf{1}^T$ with $E$, we have
$$A = b\left( -n I + E \right) \qquad B = - n b I + \frac{1+nb}{n} E$$
Note that $E^2=nE$, so that
$$A^2 = b^2 \left(n^2I-2nE+E^2\right) = -nbA \qquad \text{and}\qquad AE=0$$
By the ever-popular Sherman-Morrison identity,
$$B^{-1} = -\frac{1}{nb} I + k E$$
for some $k$. Whence,
$$AB^{-1}A = -\frac{1}{nb}A^2 + 0= A$$
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