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અહી $x =\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ અને $A =\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$ આપેલ છે. જો $k \in N$, if $X ^{\prime} A ^{ k } X =33$, હોય તો $k$ ની કિમંત મેળવો.
$99$
$100$
$23$
$10$
Solution
$X =\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] ; A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$
$X^{ T } A ^{ K } X =33$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]^{ k }\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
${\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=33 }$
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)
