If A is any square matrix of order 3 and B $=\begin{bmatrix} sin \theta & cos \theta & 0\\-cos\theta & sin \theta & 0\\0 & 0 & a\end{bmatrix};$ a is any constant, then |AB|is equal to :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $a|A|$
$|B|=a(\sin^2θ+\cos^2θ)=a$
so $|AB|=|A||B|$
$⇒a|A|$