# If $m = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]$ and $n = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right 0 votes 35 views in Matrices closed If \(m = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]$ and $n = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]$, then what is the value of the determinant of m cos θ – n sin θ?
1. -1
2. 0
3. 1
4. 2
5. None of these

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selected

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.