Let the matrix satisfy for . Then the sum of all the elements of is: [2025]

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Solution

Q.
Let the matrix $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is: [2025]

1 44

2 39

3 52

4 53

Ans.
(4)

We have, $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$\Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}]$

Since, $A^{n} = A^{n - 2} + A^{2} - I, n > 3$

$\Rightarrow A^{50} = A^{48} + A^{2} - I$

   $= [A^{46} + A^{2} - I] + (A^{2} - I)$

   $= A^{46} + 2 (A^{2} - I)$

   $= A^{44} + 3 (A^{2} - I)$

   $= A^{2} + 24 (A^{2} - I)$

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

   $= 25 [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] - 24 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{matrix}]$

So, sum of elements of $A^{50} = 53$ .

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