Computing $A^{50}$ for a given matrix $A$ $A =\left(
               \begin{array}{ccc}
                 1 & 0 & 0 \\
                 1 & 0 & 1 \\
                 0 & 1 & 0 \\
               \end{array}
             \right)$ then what would be $A^{50}$?
For real entries
 A: Let $e_1,e_2,e_3$ be the basis vectors. We have $Ae_2=e_3$ and $Ae_3=e_2$, so $A^{50}$ fixes both $e_2$ and $e_3$. We only need to check what $A^{50} e_1$ is. Calculate the first few:
$$
Ae_1=e_1+e_2,~A^2e_1 =e_1+e_2+e_3, A^3e_1=e_1+2e_2+e_3.
$$
We use induction to prove $A^{2k-1}e_1=e_1+k e_2 + (k-1)e_3$ and $A^{2k}e_1=e_1+ke_2+ke_3$.
So $A^{50}e_1=e_1+25e_2+25e_3$ and
$$
A^{50}=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\end{pmatrix}
$$
A: the characteristic polynomial is 
\[ 
  \chi_A(t) = (1-t)\bigl(t^2 - 1)
\]
so the eigenvalues are $\pm 1$, we have 
\[
   A - 1 = \begin{pmatrix} 0 & 0 & 0\\\ 1 & -1 & 1 \\\ 0 & 1 & -1 \end{pmatrix} 
\]
which has defect $1$ (so $A$ isn't diagonalizable). It holds
\[
   (A - 1)^2 = \begin{pmatrix} 0 & 0 & 0\\\ -1 & 2 & -2 \\\ 1 & -2 & 2 
 \end{pmatrix} 
\]
so a basis of $\ker (A-1)^2$ is $\\{(2,1,0), (2,0,-1)\\}$, we have $A \cdot (2,1,0) = (0, 1, 1)$, so $\\{(2,1,0), (0,1,1)\\}$ is the basis of $\ker(A-1)^2$ we will use. It holds 
\[
   A + 1 = \begin{pmatrix} 2 & 0 & 0\\\ 1 & 1 & 1 \\\ 0 & 1 & 1 \end{pmatrix} 
\]
we choose the basis $\{(0,1,-1)\}$ of its kernel. So we let 
\[
   S := \begin{pmatrix} 2 & 0 & 0\\\ 1 & 1 & 1 \\\ 0 & 1 & -1 \end{pmatrix} 
\]
Then 
\[
   S^{-1}AS = \begin{pmatrix} 1 & 0 & 0\\\ 1 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix} 
\]
so
\[
   S^{-1}A^{50}S = \begin{pmatrix} 1 & 0 & 0\\\ 50 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix} 
\]
and finally 
\[
   A^{50} = \begin{pmatrix} 1 & 0 & 0\\\ 25 & 1 & 0 \\\ 25 & 0 & 1 \end{pmatrix} 
\]
A: The characteristic polynomial of $A$ is 
$$X^3-X^2-X+1=(x-1)^2(x+1)$$
The Division Theorem tells us that 
$$X^{50}=(X^3-X^2-X+1)Q(x)+ax^2+bx+c (*)$$
Plugging in $x=1$, $x=-1$ yield 
$$1=a+b+c$$
$$1=a-b+c$$
Thus $b=0$ and $a+c=1$.
Differentiating $(*)$ and using the fact that $(X^3-X^2-X+1)Q(x)=(x-1)^2(something)$ we get
$$50x^{49}=(x-1)(junk)+2ax+b$$
Plugging in $a=1$ we get
$$50=2a+b$$
Thus $a=25, b=0, c=-24$
This proves that 
$$X^{50}=(X^3-X^2-X+1)Q(x)+25x^2-24$$
Using the fact that $A^3-A^2-A+I =0$ we get
$$A^{50}=25A^2-24I$$
P.S. This is basically the same proof as Ihf 's, the only difference is that long division can replace the induction process. 
To get the general formula,  repeating the process for
$$X^{n}=(X^3-X^2-X+1)Q(x)+ax^2+bx+c (*)$$
yields 
$$a+b+c=1$$
$$a-b+c =(-1)^n$$
$$2a+b=n$$
and
$$A^n= aA^2+bA+cI$$
A: By the Cayley–Hamilton theorem, $A^3= A^2+A-I$. From this relation we can find all powers of $A$ in terms of $A^2$, $A$, and $I$. By induction, we get
$$
A^{2n} = n A^2 -(n-1)I,
\qquad
A^{2n+1} = n A^2 +A-nI
$$
Hence, $A^{50} = 25A^2-24I$.
A: With PARI/GP:
gp >[1,0,0;1,0,1;0,1,0]^50
%1 =
[1 0 0]
[25 1 0]
[25 0 1]

