<s>[INST] Let A be a matrix then $A^{50}$ is? 
If $$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\\ \end{bmatrix}$$ then ${A^{50}}$ is?                                   

How to calculate easily? What is the trick behind it? Please tell me.
 [/INST] Use the fact that:
$$A^2={\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\\ \end{pmatrix}}=I+\begin{pmatrix}0&0&0\\1&0&0\\1&0&0\\ \end{pmatrix}\equiv I+ B$$
Also note that $B^n=0, \forall n>1$, thus using the binomial theorem, and noting that all higher powers of $B$ are dropped:
$$A^{2n}=(I+B)^n=I+nB+ {n\choose 2}B^2+\cdots+ B^n=I+nB$$
Thus, 
$$A^{50}=I+25B=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\\ \end{pmatrix}$$
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