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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.

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    How are $P $ and $A $ singular?? Please format properly using instructions [here](http://math-linux.com/latex-26/faq/latex-faq/article/how-to-write-matrices-in-latex-matrix-pmatrix-bmatrix-vmatrix-vmatrix). – 2017-01-21
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    @Rohan edited.Thanks. – 2017-01-21
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    @Moo Yes that's true – 2017-01-21
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    @Moo i am sorry for that. – 2017-01-21

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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:

$$ \require{cancel} A^n = (I+A_1)^n = I^n + \binom{n}{1} I^{n-1} A_1 + \cancel{\binom{n}{2} I^{n-2} A_1^2 + \cdots} = I + n A_1 = \begin{pmatrix} 1&n \\ 0&1 \end{pmatrix} $$

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    Thanks for your answer.But how did you get$Q^n=PA^nP^{-1}$. – 2017-01-22
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    @navinstudent By induction for example, $n=1$ is obvious, then using the associativity of matrix multiplication: $$ Q^n = Q \, Q^{n-1} = (P \,A \,P^{-1})\,(P A^{n-1} P^{-1}) = P\,A\,(P^{-1}\,P)\,A^{n-1} P^{-1} = P\,A\,A^{n-1} P^{-1} \\ = P\,A^{n} P^{-1} $$ – 2017-01-22