Practicing Success
Practicing Success
CUET Preparation Today
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 : |
$a|A|$ $a^2|A|$ a |A| |
$a|A|$ |
The correct answer is Option (1) → $a|A|$ $|B|=a(\sin^2θ+\cos^2θ)=a$ so $|AB|=|A||B|$ $⇒a|A|$ |
