If $A=\begin{bmatrix}\cos θ&\sin θ&0\\-\sin θ&\cos θ&0\\0&0&1\end{bmatrix}$ and B is a square matrix of order 3, then $|AB|$ is equal to:

$|B|$

$A=\begin{bmatrix}\cos θ&\sin θ&0\\-\sin θ&\cos θ&0\\0&0&1\end{bmatrix}$

$|AB|=|A||B|$

so |A| → expansion along Row 3

we get $1×\begin{vmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{vmatrix}=\cos^2 θ+\sin^2 θ=1(|A|)$

so $|AB|=|A||B|$

$=|B|$