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Given that $A=\begin{bmatrix} 1& 0& 0\\1 &0 &1 \\0 &1 &0 \end{bmatrix}$       

              $A^{2}=\begin{bmatrix} 1& 0& 0\\1 &1 &0 \\1 &0 &1 \end{bmatrix}$   

Chracteristics equation $|A-\lambda I|=0$

  $\begin{vmatrix} 1-\lambda& 0& 0\\1 &0-\lambda &1 \\0 &1 &0-\lambda \end{vmatrix}=0$

 $\begin{vmatrix} 1-\lambda& 0& 0\\1 &-\lambda &1 \\0 &1 &-\lambda \end{vmatrix}=0$

$(1-\lambda)[\lambda^{2}-1]-0+0=0$

$(1-\lambda)[\lambda^{2}-1^{2}]=0$

$(\lambda-1)[(\lambda-1)(\lambda+1)]=0$

$(\lambda-1)^{2}(\lambda+1)=0$

$(\lambda+1)(\lambda^{2}+1-2\lambda)=0$

$\lambda^{3}-\lambda^{2}-\lambda+1=0$

By Cayley Hamilton theorem:- Every square matrix satisfy its characteristic equation.

we can write like this $A^{3}-A^{2}-A+I=0$

                                  $A^{3}=A^{2}+A-I$

                                  $A^{4}=A^{3}+A^{2}-A$   

                                  $A^{4}=A^{2}+A-I+A^{2}-A$ 

                                   $A^{4}=2A^{2}-I$ 

Similarly we get     

                                      $A^{5}=2A^{2}+A-2I$ 

                                      $A^{6}=3A^{2}-2I$    

                                       $A^{7}=3A^{2}+A-3I$ 

                                      $A^{8}=4A^{2}-3I$ 

                                       $A^{9}=4A^{2}+A-4I$ 

                                       $A^{10}=5A^{2}-4I$ 

----------------------------------------------------------------

---------------------------------------------------------------

                                      $A^{49}=24A^{2}+A-24I$ 

                                       $A^{50}=25A^{2}-24I$ 

Now,$A^{50}=25\begin{bmatrix} 1& 0& 0\\1 &1 &0 \\1 &0 &1 \end{bmatrix}-24\begin{bmatrix} 1& 0& 0\\0 &1 &0 \\0 &0 &1 \end{bmatrix}$

     $A^{50}=\begin{bmatrix} 25& 0& 0\\25 &25 &0 \\ 25&0 &25 \end{bmatrix}-\begin{bmatrix} 24& 0& 0\\0 &24 &0 \\0 &0 &24 \end{bmatrix}$

$A^{50}=\begin{bmatrix} 1& 0& 0\\25 &1 &0 \\ 25&0 &1 \end{bmatrix}$

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