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
$A^{5}=2A^{2}+A-2I$
$A^{6}=3A^{2}-2I$
$A^{7}=3A^{2}+A-3I$
$A^{8}=4A^{2}-3I$
$A^{9}=4A^{2}+A-4I$
$A^{10}=5A^{2}-4I$
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$A^{49}=24A^{2}+A-24I$
$A^{50}=25A^{2}-24I$
Now,$A^{50}=25\begin{bmatrix} 1& 0& 0\\1 &1 &0 \\1 &0 &1 \end{bmatrix}-24\begin{bmatrix} 1& 0& 0\\0 &1 &0 \\0 &0 &1 \end{bmatrix}$
$A^{50}=\begin{bmatrix} 25& 0& 0\\25 &25 &0 \\ 25&0 &25 \end{bmatrix}-\begin{bmatrix} 24& 0& 0\\0 &24 &0 \\0 &0 &24 \end{bmatrix}$
$A^{50}=\begin{bmatrix} 1& 0& 0\\25 &1 &0 \\ 25&0 &1 \end{bmatrix}$