Let \(A=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\). If \(k, l \in \mathbb{N}\) be such that \(A^{k} B^{l}=I\), then the minimum value of \(k+l\) is ________.
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Correct Answer: 10
Solution and Explanation
\(B^2=I\), so for even l, \(B^l=I\) and the equation reduces to \(A^k=I\).
\(A^2=\frac12\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\), \(A^4=(A^2)^2=-I\), \(A^8=I\).
The smallest natural k with \(A^k=I\) is 8 (A is a rotation through -45 degrees, so \(A^k=I\) only when k is a multiple of 8).
The smallest even natural l is 2. Minimum \(k+l=8+2=10\).
Answer: 10
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