Correct Answer - Option 3 : 1
Concept:
Scalar Multiplication of Matrices
- If A = [aij] m × n is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar k ⇔ [aij]m×n = [k (aij)]m×n
- If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a11 × a22) – (a12 – a21).
- sin2 x + cos2 x = 1
Calculation:
Given:
\(m = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\) And \(n = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]\)
Now,
\({\rm{m\;cos\;\theta }} = \left[ {\begin{array}{*{20}{c}} {{\rm{cos\;\theta }}}&0\\ 0&{{\rm{cos\;\theta }}} \end{array}} \right]\) And \({\rm{n\;sin\;\theta }} = \left[ {\begin{array}{*{20}{c}} 0&{{\rm{sin\;\theta }}}\\ { - {\rm{sin\;\theta }}}&0 \end{array}} \right]\)
Now find the value of (m cos θ – n sin θ)
\(\Rightarrow {\rm{m\;cos\;\theta \;}}-{\rm{\;n\;sin\;\theta \;}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{\rm{cos\;\theta }}}&0\\ 0&{{\rm{cos\;\theta }}} \end{array}} \right] - \;\left[ {\begin{array}{*{20}{c}} 0&{{\rm{sin\;\theta }}}\\ { - {\rm{sin\;\theta }}}&0 \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} {{\rm{cos\;\theta }} - {\rm{\;}}0}&{0 - {\rm{\;sin\;\theta }}}\\ {0 - \left( { - {\rm{sin\;\theta }}} \right)}&{{\rm{cos\;\theta }} - 0} \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} {{\rm{cos\;\theta }}}&{ - {\rm{sin\;\theta }}}\\ {{\rm{sin\;\theta }}}&{{\rm{cos\;\theta }}} \end{array}} \right]\)
Determinant of m cos θ – n sin θ = (cos θ × cos θ) – (-sin θ × sin θ)
= cos2 θ + sin2 θ = 1
∴ Option 3 is correct.