Matrices Question 5
Question 5 - 25 January - Shift 2
Let $A= \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix} $ and $B= \begin{bmatrix} 1 & -i \\ 0 & 1\end{bmatrix} $, where
$i=\sqrt{-1}$. If $M=A^{\top} B A$, then the inverse of the matrix $AM^{2023} A^{T}$ is
(1) $ \begin{bmatrix} 1 & -2023 i \\ 0 & 1\end{bmatrix} $
(2) $ \begin{bmatrix} 1 & 0 \\ -2023 i & 1\end{bmatrix} $
(3) $ \begin{bmatrix} 1 & 0 \\ 2023 i & 1\end{bmatrix} $
(4) $ \begin{bmatrix} 1 & 2023 i \\ 0 & 1\end{bmatrix} $
Show Answer
Answer: (4)
Solution:
Formula: Properties of Transpose of matrix, Properties of matrix multiplication, Properties of Inverse of a matrix, Properties Of Positive Integral Powers Of A Square Matrix
$AA^{T}= \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $
$B^{2}= \begin{bmatrix} 1 & -i \\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & -i \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & -2 i \\ 0 & 1\end{bmatrix} $
$B^{3}= \begin{bmatrix} 1 & -3 i \\ 0 & 1\end{bmatrix} $
$B^{2023}= \begin{bmatrix} 1 & -2023 i \\ 0 & 1\end{bmatrix} $
$M=A^{T} BA$
$M^{2}=M \cdot M=A^{T} BA A^{T} BA=A^{T} B^{2} A$
$M^{3}=M^{2} \cdot M=A^{T} B^{2} A A^{T} B A=A^{T} B^{3} A$
$M^{2023}=$
$A^{T} B^{2023} A$
$AM^{2023} A^{T}=AA^{T} B^{2023} AA^{T}=B^{2023}$
$= \begin{bmatrix} 1 & -2023 i \\ 0 & 1\end{bmatrix} $
Inverse of $(AM^{2023} A^{T})$ is $ \begin{bmatrix} 1 & 2023 i \\ 0 & 1\end{bmatrix} $
