Show that $(A*B*A^{-1})^2 = B^2$

Question

Show that $(A*B*A^{-1})^2 = B^2$

I tried:

$(A*B*A^{-1})*(A*B*A^{-1})$ $\hspace{5mm}$ (1)

$= A*B*A^{-1}*A*B*A{-1}$ $\hspace{5mm}$ (2)

$= A*B*B*A^{-1}$ $\hspace{5mm}$ (3)

$ = A*B^2*A^{-1}$ $\hspace{5mm}$ (4)

In step $(4)$ I do not know how to clear $B ^ 2$

Answer 1

This does not hold in general.

Let $A=\left(\begin{matrix}1&1\\2&3\end{matrix}\right)$ and $B=\left(\begin{matrix}1&-1\\1&2\end{matrix}\right)$. Then $$(ABA^{-1})^2=\left[\left(\begin{matrix}1&1\\2&3\end{matrix}\right)\left(\begin{matrix}1&-1\\1&2\end{matrix}\right)\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right)\right]^2=\left(\begin{matrix}9&-3\\21&-6\end{matrix}\right)$$ whereas $$B^2=\left(\begin{matrix}1&-1\\1&2\end{matrix}\right)\left(\begin{matrix}1&-1\\1&2\end{matrix}\right)=\left(\begin{matrix}0&-3\\3&3\end{matrix}\right)$$