Question

# 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}

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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 ${{A}^{50}}$

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.