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(A) Given $A^n = A^{n-2} + A^2 - I$. For $n=50$,we have $A^{50} = A^{48} + A^2 - I$.By repeating this recurrence relation,we get $A^{50} = A^{48} + (A

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$53$

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(A) Given $A^n = A^{n-2} + A^2 - I$. For $n=50$,we have $A^{50} = A^{48} + A^2 - I$.By repeating this recurrence relation,we get $A^{50} = A^{48} + (A^2 - I) = A^{46} + 2(A^2 - I) = A^{44} + 3(A^2 - I) = \dots = A^2 + 24(A^2 - I) = 25A^2 - 24I$.First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 0 & 0 \ 1 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 1 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}$.Now,$A^{50} = 25 \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix} - 24 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 25-24 & 0 & 0 \ 25 & 25-24 & 0 \ 25 & 0 & 25-24 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 25 & 1 & 0 \ 25 & 0 & 1 \end{bmatrix}$.The sum of all elements is $1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 53$.

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

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