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.
 A: 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, 
$$A^{50}=I+25B=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\\ \end{pmatrix}$$
A: Although there are easier ways in this particular case, I will outline a trick involving the Cayley-Hamilton theorem. 
We have $\det\begin{pmatrix}
1-X & 0&0\\
1& -X& 1\\
0&1&-X
\end{pmatrix}=-(X+1)(1-X)^2=Q(X)$. Hence there are two eigenvalues, namely $-1$ and $1$. We know that $A$ is diagonalizable if there are two independent eigenvectors belonging to the eigenvalue $1$, but that's not the case here. 
By the Cayley-Hamilton theorem, we have that $-(A+1)(1-A)^2=0$. Now consider the polynomial $P(X)=X^{50}$. We can divide this polynomial by $Q(X)$ and obtain $$P(X)=S(X)Q(X)+R(X)$$
where $R(X)$ is a polynomial of degree at most $2$. Notice that $P(A)=R(A)$ by the Cayley-Hamilton theorem. Write $R(X)=a+bX+cX^2$. Now notice that $$50X^{49}=P'(X)=S'(X)Q(X)+S(X)Q'(X)+R'(X).$$
It follows that $50=P'(1)=R'(1)=b+2c$, here I used that $1$ is a double root and hence $Q'(1)=0$. Also, $1=P(1)=R(1)=a+b+c$ and $P(-1)=R(-1)=a-b+c$. Now you can solve for $a,b$ and $c$. You should find that $a=-24,b=0$ and $c=25$. It follows that $$A^{50}=R(A)=-24I+25A^2=-24I+25\begin{pmatrix}
1&0&0\\
1&1&0\\
1&0&1
\end{pmatrix}=\begin{pmatrix}
1&0&0\\
25&1&0\\
25&0&1
\end{pmatrix}.$$
A: Show that $A^{2^n}=\begin{bmatrix} 1 & 0 & 0 \\ 2^{n-1} & 1 & 0 \\ 2^{n-1} & 0 & 1 \end{bmatrix}$ using mathematical induction and calculate $A^{50}=A^{32}A^{16}A^2$.
A: Hint
$$A^2={\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\\ \end{bmatrix}}$$
Observe that 
$$A^2 =I_3+B $$
Where B is nilpotent...$B^2=0_3$
$$A^{50}=(A^2)^{25}=(I_3+B)^{25}$$
$$............$$
A: This answer is not really useful in this case, because $A$ is not diagonalizable. Why? 
Well, the eigenvalues of $A$ are $\lambda_1 = -1$ and $\lambda_2 = 1$.
The algebraic multiplicity of $\lambda_1 = -1$ is $1$, and that of $\lambda_2 = 1$ is $2$, but the geometric multiplicity of $\lambda_2 = 1$ is $1$, which means that the union of the eigenspaces has dimension $2$, not $3$ ($A$ is a $3 \times 3$ matrix) and hence not diagonalizable.
Now, if $A$ were diagonalizable, you could write it as $A = PDP^{-1}$, where $D = \text{diagon}(\lambda_1, \lambda_2, \lambda_3)$.
It would follow that  $A^n =AA...A= (PDP^{-1})(PDP^{-1})...(PDP^{-1})$ (the matrix is multiplied $n$ times)
And hence,
$$A^{n} = PD^{n}P^{-1}$$
Note that $$D = \text{diagon}(\lambda_1, \lambda_2, \lambda_3) \Rightarrow D^{n} = \text{diagon}(\lambda_1^{n}, \lambda_2^{n}, \lambda_3^{n})$$
This gives an easy formula for $D^{n}$, and then you just have to multiply it with $P$ and $P^{-1}$ to obtain $A^{n}$.
