CUET · MATHS · PYQ PAPER 2023
If \(A=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\) and B is a square matrix of order 3, then |AB| is equal to:
- A \(|B|^2\)
- B \(|B|\)
- C \(\sin ^2 \theta|B|\)
- D \(\cos ^2 \theta|B|\)
Answer & Solution
Correct Answer
(B) \(|B|\)
Step-by-step Solution
Detailed explanation
\(|A| = \cos \theta (\cos \theta \cdot 1 - 0 \cdot 0) - \sin \theta (-\sin \theta \cdot 1 - 0 \cdot 0) + 0\) \(|A| = \cos^2 \theta + \sin^2 \theta = 1\) \(|AB| = |A||B|\) \(|AB| = 1 \cdot |B| = |B|\)
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