JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 16)
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 $$\mathrm{k} \in N$$, if $$X^{\prime} A^{k} X=33$$, then $$\mathrm{k}$$ is equal to _______.
Answer
10
Explanation
Given $A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$
$A^{2}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad A^{4}=\left[\begin{array}{ccc}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A^{k}=\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad X^{\prime} A^{k} X=[111]\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 k+3]$
$\Rightarrow[3 k+3]=33$ (here it shall be [33] as matrix can't be equal to a scalar)
i.e. $[3 k+3]=33$
$3 k+3=[33] \quad \Rightarrow k=10$
If $k$ is odd and apply above process, we don't get odd value of $k$
$\therefore k=10$
$A^{2}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad A^{4}=\left[\begin{array}{ccc}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A^{k}=\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad X^{\prime} A^{k} X=[111]\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 k+3]$
$\Rightarrow[3 k+3]=33$ (here it shall be [33] as matrix can't be equal to a scalar)
i.e. $[3 k+3]=33$
$3 k+3=[33] \quad \Rightarrow k=10$
If $k$ is odd and apply above process, we don't get odd value of $k$
$\therefore k=10$
Comments (0)
