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 $k \in \mathbb{N}$, if $x^{T}A^{k}x=33$, then $k$ is equal to.

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Aicharya · Accepted Answer

Given, $X = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1 \end{bmatrix}$ Also given, $X^T A^K X = 33$ Now putting the value of matrices in $X^T A^K X = 33$ we get, $\begin{bmatrix}1 & 1 & 1\end{bmatrix} \begin{bmatrix}-1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1\end{bmatrix}^k \begin{bmatrix}1 \ 1 \ 1\end{bmatrix} = 33$ Now finding $A^2 = \begin{bmatrix}-1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1\end{bmatrix} \begin{bmatrix}-1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1\end{bmatrix} = \begin{bmatrix}1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$ And $A^4 = \begin{bmatrix}1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 12 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$ Similarly $A^8 = \begin{bmatrix}1 & 0 & 24 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$ And $A^{10} = \begin{bmatrix}1 & 0 & 6 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 24 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 30 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$ So, for $K ightarrow$ Even $A^K = \begin{bmatrix}1 & 0 & 3K \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix}$ Now again putting the value in $X^T A^K X = 33$ we get, $\Rightarrow \begin{bmatrix}1 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 3K \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 \ 1 \ 1\end{bmatrix} = 33$ $\Rightarrow \begin{bmatrix}1 & 1 & 3K+1\end{bmatrix} \begin{bmatrix}1 \ 1 \ 1\end{bmatrix} = 33$ $\Rightarrow \begin{bmatrix}3K+3\end{bmatrix} = 33$ Now assuming $33$ as $\begin{bmatrix}33\end{bmatrix}$ We get, $3K + 3 = 33 \Rightarrow K = 10$ Now, if $K$ is odd $X^T A^K X = 33$ We can rewrite above expression as $X^T A A^{K-1} X = 33$ $\Rightarrow \begin{bmatrix}1 & 1 & 1\end{bmatrix} \begin{bmatrix}-1 & 2 & 3 \ 0 & 1 & 6 \ 0 & 0 & -1\end{bmatrix} \begin{bmatrix}1 & 0 & 3K-3 \ 0 & 1 & 0 \ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 \ 1 \ 1\end{bmatrix} = 33$ $\Rightarrow \begin{bmatrix}-1 & 3 & 8\end{bmatrix} \begin{bmatrix}3K-2 \ 1 \ 1\end{bmatrix} = \begin{bmatrix}33\end{bmatrix}$

Let $x=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $A=\begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 &…

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