Let x = [ array ll 1 1 1 array ] and A = [ cases - 1 & 2 & 3 0 & 1...

Given,

X = \left[$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right] \text{ and } A = \left[\begin{cases}- 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1\end{cases}\right]

Also given, ${X}^{T}{A}^{K}X = 33$

Now putting the value of matrices in ${X}^{T}{A}^{K}X = 33$ we get,

\left[$\begin{cases}1 & 1 & 1\end{cases}$\right]{\left[$\begin{cases}- 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1\end{cases}$\right]}^{k}\left[$$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$$\right] = 33$</p><p>Now finding${A}^{2} = \left[$\begin{cases}- 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1\end{cases}$\right]\left[$\begin{cases}- 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>And${A}^{4} = \left[$\begin{cases}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]\left[$\begin{cases}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>Similary${A}^{8} = \left[$\begin{cases}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>And${A}^{10} = \left[$\begin{cases}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]\left[$\begin{cases}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 30 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>So, for$K \to$Even${A}^{K} = \left[$\begin{cases}1 & 0 & 3K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>Now again putting the value in${X}^{T}{A}^{K}X = 33$we get,</p><p>$⇒ \left[$\begin{cases}1 & 1 & 1\end{cases}$\right]\left[$\begin{cases}1 & 0 & 3K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]\left[$$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$$\right] = 33$</p><p>$⇒ \left[$\begin{cases}1 & 1 & 3K + 1\end{cases}$\right]\left[$$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$$\right] = 33$</p><p>$⇒ \left[3K + 3\right] = 33$</p><p>Now assuming$33$as \left[33\right]$

We get, $3K + 3 = 33 ⇒ K = 10$

Now, if $K$ is odd ${X}^{T}{A}^{K}X = 33$

We can rewrite above expression as ${X}^{T}A{A}^{K - 1}X = 33$

⇒ \left[\begin{cases}1 & 1 & 1\end{cases}\right]\left[\begin{cases}- 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1\end{cases}\right]\left[\begin{cases}1 & 0 & 3k - 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right]\left[$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right] = 33

⇒ \left[\begin{cases}- 1 & 3 & 8\end{cases}\right]\left[$\begin{array}{ll}3k - 2 \\ 1 \\ 1\end{array}$\right] = \left[33\right]

$⇒ \left[- 3k + 13\right] = \left[33\right]$

$⇒ k = 20 / 3$ (not possible)

So, $k = 10$ is the required answer.