Let x = [ l 1 1 1 ] and A = [ ccc -1 & 2 & 3 0 & 1 & 6 0 & 0 & -1 ]. For k N, if X ^ A ^ k X =33, then k is equal to.

Let x = [ l 1 1 1 ] and A = [ ccc -1 & 2 & 3 0 & 1 & 6 0 & 0 & -1…

MCQ

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 $k \in N,$ if $X ^{\prime} A ^{ k } X =33,$ then $k$ is equal to.

A
$99$
B
$100$
C
$23$
✓
$10$

✓

Answer

Correct option: D.

$10$

$X =\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] ; A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]X^{ T } A ^{ K } X =33$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]^{ k }\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
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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