$X^{ T } A ^{ K } X =33$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]^{ k }\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
As $A^{2}=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A ^{4}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A ^{8}=\left[\begin{array}{llll}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{10}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 30 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
for $K \rightarrow$ Even $A ^{ K }=\left[\begin{array}{ccc}1 & 0 & 3 K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$X ^{ T } A ^{ K } X =33$ (This is not correct)
1] $\left[\begin{array}{ccc}1 & 0 & 3 K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]$
$X ^{ T } A ^{ K } X =33$ (This is not correct)
$\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 3 K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 1 & 3 K +1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 K +3]$
$\therefore 3 K +3=33 \therefore K =10$
$\therefore 3 K +3=33 \therefore K =10$
But it should be dropped as 33 is not matrix
If $K$ is odd
$\begin{array}{l}X^{ T } A^{ K } X =33 \\X ^{ T } AA ^{ K -1} X =33\end{array}$
$\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 3 k-3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33$
${\left[\begin{array}{lll}-1 & 3 & 8\end{array}\right]\left[\begin{array}{l}3 k -2 \\ 1 \\ 1\end{array}\right]=[33] }$
${[-3 k +13]=[33] }$
$k =20 / 3$ (not possible)
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Statement $-2 :$ $\;\mathop \smallint \limits_a^b {\rm{f}}\left( {\rm{x}} \right)dx = \mathop \smallint \limits_a^b {\rm{f}}\left( {a + b - x} \right)\;dx$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is