If \(\rm m = \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right] and \ n= \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\) then what is the value of the determinant of m sinθ + n cosθ 

Calculation: 

m sinθ + n cosθ = sin θ \(\rm \left[ \begin{array}{cc} 2 &0\\ 0&1\end{array}\right]\) + cos θ \(\rm \left[ \begin{array}{cc} 0 &1\\ -2&0\end{array}\right]\)

= \(\rm \left[ \begin{array}{cc} 2\sin θ &0\\ 0&\sin θ\end{array}\right]\)+\(\rm \left[ \begin{array}{cc} 0 &\cosθ\\ -2\cosθ&0\end{array}\right]\)

= \(\rm \left[\begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right]\)

Now,

|m sinθ + n cosθ| = \(\rm \left| \begin{array}{cc} 2\sin θ &\cos θ\\ -2\cos θ&\sin θ\end{array}\right|\)

= \(2\sin^2\theta \) + \(\rm 2 cos^2\theta \)

= 2(\(\sin ^2 \theta +\cos^2\theta \))

= 2

Hence, option (4) is correct.