JEE Main 2022 — Mathematics Algebra

Given,

X=[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]&A=[\begin{matrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{matrix}]

Also given, ${X}^{T}{A}^{K}X=33$XTAKX=33

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

$$[\begin{matrix}1 & 1 & 1\end{matrix}]{[\begin{matrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{matrix}]}^{k}[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=33$$[1​1​1​][−100​210​36−1​]k[111​]=33

Now finding ${A}^{2}=[\begin{matrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{matrix}][\begin{matrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{matrix}]=[\begin{matrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$A2=[−100​210​36−1​][−100​210​36−1​]=[100​010​601​]

And ${A}^{4}=[\begin{matrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]=[\begin{matrix}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$A4=[100​010​601​][100​010​601​]=[100​010​1201​]

Similary ${A}^{8}=[\begin{matrix}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$A8=[100​010​2401​]

And ${A}^{10}=[\begin{matrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]=[\begin{matrix}1 & 0 & 30 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$A10=[100​010​601​][100​010​2401​]=[100​010​3001​]

So, for $K\rightarrow$K→ Even ${A}^{K}=[\begin{matrix}1 & 0 & 3K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$AK=[100​010​3K01​]

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

$$\Rightarrow [\begin{matrix}1 & 1 & 1\end{matrix}][\begin{matrix}1 & 0 & 3K \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=33$$⇒[1​1​1​][100​010​3K01​][111​]=33

$$\Rightarrow [\begin{matrix}1 & 1 & 3K+1\end{matrix}][\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=33$$⇒[1​1​3K+1​][111​]=33

$$\Rightarrow [3K+3]=33$$⇒[3K+3]=33

Now assuming $33$33 as $[33]$[33]

We get, $3K+3=33\Rightarrow K=10$3K+3=33⇒K=10

Now, if $K$K is odd ${X}^{T}{A}^{K}X=33$XTAKX=33

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

$$\Rightarrow [\begin{matrix}1 & 1 & 1\end{matrix}][\begin{matrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{matrix}][\begin{matrix}1 & 0 & 3k-3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=33$$⇒[1​1​1​][−100​210​36−1​][100​010​3k−301​][111​]=33

$$\Rightarrow [\begin{matrix}-1 & 3 & 8\end{matrix}][\begin{matrix}3k-2 \\ 1 \\ 1\end{matrix}]=[33]$$⇒[−1​3​8​][3k−211​]=[33]

$$\Rightarrow [-3k+13]=[33]$$⇒[−3k+13]=[33]

$\Rightarrow k=20/3$⇒k=20/3 (not possible)

So, $k=10$k=10 is the required answer.