Mathematics - Matrices Question with Solution | TestHub

Given,

$X=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right] \& A=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]$

Also given, $X^T{A}^{K}X=33$

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

$\left[\begin{array}{lll}1& 1& 1\end{array}\right]{\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]}^{\mathrm{k}}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=33$

Now finding   $A^2=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

And $A^4=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 12\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Similary $A^8=\left[\begin{array}{lll}1& 0& 24\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

And $A^{10}=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 0& 24\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 30\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

So, for $K\to$ Even $A^K=\left[\begin{array}{ccc}1& 0& 3K\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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

$\Rightarrow \left[\begin{array}{lll}1& 1& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 3 K\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=33$

$\Rightarrow \left[\begin{array}{lll}1& 1& 3 K+1\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=33$

$\Rightarrow \left[3 K+3\right]=33$

Now assuming $33$ as $\left[33\right]$

We get, $3 K+3=33 \Rightarrow K=10$

Now, if $K$ is odd $X^T{A}^{K}X=33$

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

$\Rightarrow \left[\begin{array}{lll}1& 1& 1\end{array}\right]\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 3k-3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=33$

$\Rightarrow \left[\begin{array}{lll}-1& 3& 8\end{array}\right]\left[\begin{array}{l}3k-2\\ 1\\ 1\end{array}\right]=\left[33\right]$

$\Rightarrow \left[-3\mathrm{k}+13\right]=\left[33\right]$

$\Rightarrow \mathrm{k}=20/3$ (not possible)

So, $k=10$ is the required answer.