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If \[A=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right)\] then $ {{A}^{50}} $ equals
\[\begin{align}
  & A.\left( \begin{matrix}
   1 & 0 & 0 \\
   25 & 1 & 0 \\
   25 & 0 & 0 \\
\end{matrix} \right) \\
 & B.{{A}^{2}}+24\left( {{A}^{2}}-I \right) \\
 & C.\left( \begin{matrix}
   1 & 25 & 25 \\
   1 & 0 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & D.\text{ None of these} \\
\end{align}\]

Answer
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Hint: The value of matrix $ {{A}^{50}} $ cannot be directly calculated just by simply multiplying A 50 times. Using hit and trial method find $ {{A}^{2}},{{A}^{3}}.... $ until you get some pattern. Once you get a pattern of all elements in the matrices, predict the matrix $ {{A}^{50}} $

Complete step-by-step answer:
Given matrix A is a $ 3\times 3 $ matrix having 3 rows and 3 columns.
\[A=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right)\left( 3\times 3 \right)\]
We need to compute $ {{A}^{50}} $ if it was given to calculate $ {{A}^{2}}\Rightarrow {{A}^{3}} $ we could have just multiplied $ A\times A $ and $ A\times A\times A $ respectively but here $ {{A}^{50}} $ is to be calculated, so, simply multiplication of $ A\times A $ up to 50 times is not possible.
We can use hit on trial method which is just a simple method.
\[A=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right)\]
Now, find $ A\times A\text{ i}\text{.e}\text{. }{{\text{A}}^{\text{2}}} $ multiplying given matrix with itself.
By multiplication property, two matrices can only be multiplied if columns of first matrix is equal to rows of second matrix. Here, \[A=3\times 3\text{ matrix}\] so when A will be multiplied with itself it would satisfy the multiplication condition.
Multiplying $ A\times A $
\[A\times A=\begin{matrix}
   \left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right) \\
   {{1}^{st}} \\
\end{matrix}\times \begin{matrix}
   \left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right) \\
   {{2}^{nd}} \\
\end{matrix}\]
Elements of row of first matrix will get multiplied with all the elements of all the column of second matrix and will get add up.
\[\begin{align}
  & A\times A=\left( \begin{matrix}
   1\times 1+0\times 1+0\times 0 & 1\times 0+0\times 0+0\times 1 & 1\times 0+0\times 1+0\times 0 \\
   1\times 1+0\times 1+1\times 0 & 1\times 0+0\times 0+1\times 1 & 1\times 0+0\times 1+1\times 0 \\
   0\times 1+1\times 1-0\times 0 & 0\times 0+1\times 0-0\times 1 & 0\times 0+1\times 1-0\times 0 \\
\end{matrix} \right) \\
 & {{A}^{2}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 1 & 0 \\
   1 & 0 & 1 \\
\end{matrix} \right) \\
\end{align}\]
Similarly, $ {{A}^{3}} $ would be:
\[\begin{align}
  & {{A}^{2}}\times A=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 1 & 0 \\
   1 & 0 & 1 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right) \\
 & {{A}^{2}}\times A=\left( \begin{matrix}
   1\times 1+0\times 3+0\times 0 & 1\times 0+0\times 0+0\times 1 & 1\times 0+0\times 1+0\times 0 \\
   1\times 1+1\times 1+0\times 0 & 1\times 0+1\times 0+0\times 1 & 1\times 0+1\times 1+0\times 0 \\
   1\times 1+1\times 1+1\times 0 & 1\times 0+1\times 0+1\times 0 & 1\times 0+1\times 1+1\times 0 \\
\end{matrix} \right) \\
 & {{A}^{3}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 0 & 1 \\
   2 & 0 & 1 \\
\end{matrix} \right) \\
\end{align}\]
Again find $ {{A}^{4}}={{A}^{3}}\times A $
\[{{A}^{3}}\times A=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 0 & 1 \\
   1 & 1 & 0 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right)\]
After multiplying we get \[{{A}^{4}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   2 & 0 & 1 \\
\end{matrix} \right)\]
We have:
\[{{A}^{1}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 0 & 1 \\
   0 & 1 & 0 \\
\end{matrix} \right);{{A}^{2}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   1 & 1 & 0 \\
   1 & 0 & 1 \\
\end{matrix} \right);{{A}^{3}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 0 & 1 \\
   1 & 1 & 0 \\
\end{matrix} \right);{{A}^{4}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   2 & 0 & 1 \\
\end{matrix} \right)\]
Can you observe the pattern in $ {{A}^{1}}\leftrightarrow {{A}^{3}} $ with odd powers and $ {{A}^{2}}\leftrightarrow {{A}^{4}} $ with even powers.
First row always remain = 1, 0, 0. Observe column first, second and third elements are getting increased by 1 in A with even powers $ {{A}^{2}}\leftrightarrow {{A}^{4}} $ So, by hit and trial we see $ {{A}^{50}} $ as:
 $ {{A}^{50}} $ is with even power, will follow $ {{A}^{2}},{{A}^{4}} $ type
First row = 1, 0, 0
First column = Second and third elements
\[\begin{align}
  & \therefore {{A}^{2}}\leftrightarrow {{A}^{4}}:\text{ From }\begin{matrix}
   \text{1}\to \text{2} \\
   \text{1}\to \text{2} \\
\end{matrix} \\
 & \Rightarrow {{A}^{4}}\leftrightarrow {{A}^{6}}:\text{ From }\begin{matrix}
   2\to 3 \\
   2\to 3 \\
\end{matrix} \\
 & \Rightarrow {{A}^{6}}\leftrightarrow {{A}^{8}}:\text{ From }\begin{matrix}
   3\to 4 \\
   3\to 4 \\
\end{matrix} \\
\end{align}\]
Thus, $ {{A}^{2}} $ contains second and third elements of first column as $ \dfrac{1}{1} $
\[{{A}^{4}}:\begin{matrix}
   2 \\
   2 \\
\end{matrix};{{A}^{6}}:\begin{matrix}
   3 \\
   3 \\
\end{matrix};{{A}^{8}}:\begin{matrix}
   4 \\
   4 \\
\end{matrix}\]
Following this trend second and third elements of first column of $ {{A}^{50}} $ would be:
\[{{A}^{50}}:\begin{matrix}
   25 \\
   25 \\
\end{matrix}\]
By hit and trial matrix $ {{A}^{50}} $ should be:
\[{{A}^{50}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   25 & 1 & 0 \\
   25 & 0 & 1 \\
\end{matrix} \right)\]
So, the correct answer is “Option A”.

Note: This problem can also be solved by the knowledge of Eigenvalues which are $ \pm 1 $ and thus, knowing the characteristic polynomial.
\[{{x}_{A}}\left( t \right)=\left( 1-t \right)\left( {{t}^{2}}-1 \right)\]
This method is a little complex.