Question

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

• A. $\left[\begin{array}{cc}1& 2^{2021}\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 4042\\ 0& 1\end{array}\right]$

Answer 1

Answer: (D)

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