Question

If \({\rm{\;A}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]\) and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\), then what is the value of the determinant of A cos θ – B sin θ?
1. - 1
2. 0
3. 1
4. 2
5. 4

Answer 1

Correct Answer - Option 3 : 1

Concept:

Scalar Multiplication of Matrices

• If A = [a_ij] _{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 ⇔ [a_ij]_m×n = [k (a_ij)]_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:

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]\) And \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

Now,

\({\rm{A\;cos\;\theta }} = \left[ {\begin{array}{*{20}{c}} 0&{{\rm{cos\;\theta }}}\\ { - {\rm{cos\;\theta }}}&0 \end{array}} \right]\) And  \({\rm{B\;sin\;\theta }} = \left[ {\begin{array}{*{20}{c}} {{\rm{sin\;\theta }}}&0\\ 0&{{\rm{sin\;\theta }}} \end{array}} \right]\)

Now find the value of (A cos θ – B sin θ)

⇒\({\rm{\;A\;cos\;\theta \;}}-{\rm{\;B\;sin\;\theta \;}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 0&{{\rm{cos\;\theta }}}\\ { - {\rm{cos\;\theta }}}&0 \end{array}} \right] - \;\left[ {\begin{array}{*{20}{c}} {{\rm{sin\;\theta }}}&0\\ 0&{{\rm{sin\;\theta }}} \end{array}} \right]\)

=\({\rm{}}\left[ {\begin{array}{*{20}{c}} {0 - {\rm{\;sin\;\theta }}}&{{\rm{cos\;\theta }} - {\rm{\;}}0}\\ { - {\rm{cos\;\theta }} - 0}&{0 - {\rm{\;sin\;\theta }}} \end{array}} \right]\)

= \({\rm{}}\left[ {\begin{array}{*{20}{c}} { - {\rm{sin\;\theta }}}&{{\rm{cos\;\theta }}}\\ { - {\rm{cos\;\theta }}}&{ - {\rm{sin\;\theta }}} \end{array}} \right]\)

Now, Determinant of A cos θ – B sin θ

⇒ | A cos θ – B sin θ | = (-sin θ × -sin θ) – (cos θ × -cos θ)

= sin² θ + cos² θ = 1

∴ Option 3 is correct.