# Prove that $B$ is non singular and that $AB^{-1}A=A$

$$A_{n\times n}=\begin{bmatrix}a & b & b & b &. &.&.&&b\\b & a &b&b&.&.&.&&b\\b & b &a&b&.&.&.&&b\\b & . &.&.&.&.&.&&b\\b & . &.&.&.&.&.&&b\\b & b &b&b&.&.&.&&a\end{bmatrix}\text{ where } a+(n-1)b =0$$

Define $l^t=\begin{bmatrix}1&1&1&1&1&....1\end{bmatrix}$ Where $l$ is a $n\times1$ vector, and:

$$B= A+ \frac{l\cdot l^t}{n}$$

Prove that $B$ is non singular and that $AB^{-1}A=A$

What i did: