Let $$X=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$$ and $$A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$$. For $\mathrm{k} \in N$, if $X^{\prime} A^{k} X=33$, then $\mathrm{k}$ is equal to _______.

Given $A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$ <br/><br/> $A^{2}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad A^{4}=\left[\begin{array}{ccc}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ <br/><br/> $\Rightarrow A^{k}=\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ <br/><br/> $\therefore \quad X^{\prime} A^{k} X=[111]\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 k+3]$ <br/><br/> $\Rightarrow[3 k+3]=33$ (here it shall be [33] as matrix can't be equal to a scalar) <br/><br/> i.e. $[3 k+3]=33$ <br/><br/> $3 k+3=[33] \quad \Rightarrow k=10$ <br/><br/> If $k$ is odd and apply above process, we don't get odd value of $k$ <br/><br/> $\therefore k=10$