1. Tardigrade
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  3. Mathematics
  4. Let B=[1 2 0 1] and A be a 2 × 2 matrix satisfying (AT)-1= A. If X=A B AT, then AT X2021 A=

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Q. Let $B=\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$ and $A$ be a $2 \times 2$ matrix satisfying $\left(A^T\right)^{-1}=$ A. If $X=A B A^T$, then $A^T X^{2021} A=$

TS EAMCET 2021

Solution:

$ \left(A^T\right)^{-1}=A$
$ \Rightarrow \quad A^T\left(A^T\right)^{-1}=A^T \cdot A=I$
$ A^T X A=A^T\left(A B A^T\right) A $
$ =\left(A^T A\right) B\left(A^T A\right)=B $
$ A^T X^2 A=A^T\left(A B A^T\right)\left(A B A^T\right) A $
$ =\left(A^T A\right) B\left(A^T A\right) B\left(A^T A\right)=B^2$
Similarly, $A^T X^{2021} A=B^{2021}$
$B=\begin{bmatrix}1&2\\ 0&1\end{bmatrix}B^{2}=\begin{bmatrix}1&2\\ 0&1\end{bmatrix}\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}1&4\\ 0&1\end{bmatrix} $
$B^{3} = \begin{bmatrix}1&4\\ 0&1\end{bmatrix}\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}1&6\\ 0&1\end{bmatrix} $
$B^{4}=\begin{bmatrix}1&6\\ 1&1\end{bmatrix}\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}1&8\\ 0&1\end{bmatrix} $
$B^{n}=\begin{bmatrix}1&2n\\ 0&1\end{bmatrix}$
$\therefore B^{2021}=\begin{bmatrix}1&4042\\ 0&1\end{bmatrix}$