Matrices & Determinants
Orthogonal matrices, matrix powers, inverse
nta_pyq_2023_jan
Grade None
Question:
Let $A = \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -i \\ 0 & 1 \end{pmatrix}$, where $i = \sqrt{-1}$. If $M = A^T B A$, then the inverse of the matrix $AM^{2023}A^T$ is
\begin{pmatrix} 1 & -2023i \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ -2023i & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 2023i & 1 \end{pmatrix}
\begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix}
Step-by-Step Solution
Key Concept: A is orthogonal ($AA^T = I$), so $M^k = A^T B^k A$ and $AM^{2023}A^T = B^{2023}$; find $B^{2023}$ by pattern
$AA^T = I$ (A is orthogonal). $M = A^TBA$, $M^2 = A^TB^2A$, ..., $M^{2023} = A^TB^{2023}A$. $B^n = \begin{pmatrix}1 & -ni \\ 0 & 1\end{pmatrix}$, so $B^{2023} = \begin{pmatrix}1 & -2023i \\ 0 & 1\end{pmatrix}$. Thus $AM^{2023}A^T = B^{2023}$. Inverse of $B^{2023} = \begin{pmatrix}1 & 2023i \\ 0 & 1\end{pmatrix}$. Answer: (4)
Correct Answer: $\begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix}$