Multiplication of matrices --$P^{-1}Q^{n}P^{-1}$ If $P$ be a square matrix - $\begin{pmatrix} √3/2 &1/2 \\ 1/2&√3/2 \end{pmatrix}$
and $A$  be another square matrix -$\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}$. Let $Q$ be the matrix given by $Q=PAP^{-1}$ then the question is to find the value of $P^{-1}(Q^{n})P^{-1}$ for some natural number $n$. Here $P^{-1}$ represents inverse of matrix $P$.
Since $A$ have determinant unity, its inverse matrices are given by the adjoints. I calculated $Q$ and then took its powers but could not see a pattern from which $ Q^n$ can be predicted. I tried many times but failed. Please help me out. Thanks.

Hint: $Q^n = P A^n P^{-1}$ so the problem reduces to calculating $A^n$. Write:
$$
A = \begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} +  \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}=I+A_1
$$
Note that $I,A_1$ commute, $I^k=I$ and $A_1^2=0$ so by binomial expansion:
$$
\require{cancel}
A^n = (I+A_1)^n = I^n + \binom{n}{1} I^{n-1} A_1 + \cancel{\binom{n}{2} I^{n-2} A_1^2 + \cdots} = I + n A_1 = \begin{pmatrix} 1&n \\ 0&1 \end{pmatrix}
$$
