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(D) Given $X = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1 \end{bmatrix}$.First,calculat

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$10$

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(D) Given $X = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1 \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} -1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Now,calculate $A^{4} = A^{2} \cdot A^{2} = \begin{bmatrix} 1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 12 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.By induction,for any even $k$,$A^{k} = \begin{bmatrix} 1 & 0 & 3k \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.We are given $X^{T} A^{k} X = 33$. Substituting the expression for $A^{k}$:$\begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 3k \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = 33$$\begin{bmatrix} 1 & 1 & 3k+1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = 33$$1 + 1 + 3k + 1 = 33$$3k + 3 = 33$$3k = 30 \implies k = 10$.Since $10$ is an even number,this solution is valid.

Let $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}$. For $k \in N$,if $X^{T} A^{k} X = 33$,then $k$ is equal to:

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