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Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{array}\right]$ and $B =\left[\begin{array}{rr}1 & - i \\ 0 & 1\end{array}\right]$, where $i =\sqrt{-1}$. If $M = A ^{ T } BA$, then the inverse of the matrix $AM ^{2023} A ^{ T }$ is $.........$
$\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]$
$\left[\begin{array}{ll}1 & 0 \\ -2023 i & 1\end{array}\right]$
$\left[\begin{array}{ll}1 & 0 \\ 2023 i & 1\end{array}\right]$
$\left[\begin{array}{cc}1 & 2023 i \\ 0 & 1\end{array}\right]$
Solution
$AA ^{ T }=\left[\begin{array}{cc}\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$B^2=\left[\begin{array}{cc}1 & -i \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -i \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -2 i \\ 0 & 1\end{array}\right]$
$B ^3=\left[\begin{array}{cc}1 & -3 i \\ 0 & 1\end{array}\right]$
$B ^{2023}=\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]$
$M = A ^{ T } BA$
$M ^2= M \cdot M = A ^{ T } BA A ^{ T } BA = A ^{ T } B ^2 A$
$M ^3= M ^2 \cdot M = A ^{ T } B ^2 A A ^{ T } BA = A ^{ T } B ^3 A$
$M ^{2023}=$ $A ^{ T } B ^{2023} A$
$AM ^{2023} A ^{ T }= AA ^{ T } B ^{2023} AA ^{ T }= B ^{2023}$
$=\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]$
Inverse of $\left( AM ^{2023} A ^{ T }\right)$ is $\left[\begin{array}{cc}1 & 2023 i \\ 0 & 1\end{array}\right]$
