1. Tardigrade
  2. Question
  3. Mathematics
  4. Let x =[1 1 1] and A =[-1 2 3 0 1 6 0 0 -1]. For k ∈ N if X prime Ak X=33, then k is equal to:

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Q. Let $x =\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ and $A =\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}$. For $k \in N$ if $X^{\prime} A^k X=33$, then $k$ is equal to:

JEE MainJEE Main 2022Matrices

Solution:

Allen (Dropped or 10)
$ X =\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} ; A =\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}$
$ X ^{ T } A ^{ K } X =33$
$ \begin{bmatrix}1 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}^{ k }\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}=33 $
$\begin{bmatrix}1 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}=33 $
$ \text { As } A^2=\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}=\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} $
$ A^4=\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$ A^8=\begin{bmatrix}1 & 0 & 24 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
$ A^{10}=\begin{bmatrix}1 & 0 & 6 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 24 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 30 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
$ \text { for } K \rightarrow \text { Even } A^{ K }=\begin{bmatrix}1 & 0 & 3 K \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
$X ^{ T } A ^{ K } X =33$ (This is not correct)
$ \begin{bmatrix}1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 3 K \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 \\1 \\1\end{bmatrix} $
$ =\begin{bmatrix}1 & 1 & 3 K +1\end{bmatrix}\begin{bmatrix}1 \\1 \\1\end{bmatrix}=[3 K +3] $
$ \therefore 3 K +3=33 \therefore K =10 $
But it should be dropped as 33 is not matrix
If $K$ is odd
$X ^{ T } A ^{ K } X =33$
$ X ^{ T } AA { }^{ K -1} X =33$
$\begin{bmatrix}1 & 1 & 1\end{bmatrix}\begin{bmatrix}
-1 & 2 & 3 \\0 & 1 & 6 \\0 & 0 & -1\end{bmatrix}\begin{bmatrix}1 & 0& 3 k -3 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 \\1 \\1\end{bmatrix}=33 $
$\begin{bmatrix}-1 & 3 & 8\end{bmatrix}\begin{bmatrix}3 k -2 \\ 1 \\ 1\end{bmatrix}=[33]$
$ {[-3 k +13]=[33]}$
$ k =20 / 3$ (not possible)