What will be the value of \(\begin{vmatrix}cos^2⁡ θ & cosθ \, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)?(a) -1(b) 0(c) 1(d) 2The question was asked in an interview.Question is taken from Determinant topic in chapter Determinants of Mathematics – Class 12

What will be the value of \(\begin{vmatrix}cos^2⁡ θ & cosθ \, sin�

1.	What will be the value of \(\begin{vmatrix}cos^2⁡ θ & cosθ \, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)?(a) -1(b) 0(c) 1(d) 2The question was asked in an interview.Question is taken from Determinant topic in chapter Determinants of Mathematics – Class 12
Answer» Right answer is (c) 1 The explanation: The GIVEN matrix is, \(\begin{vmatrix}cos^2 θ & cosθ\, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡ θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\) Now, performing the row OPERATIONS R1 = R1 + sinθR3 and R2 = R2 –cosθR3 =\(\begin{vmatrix}cos^2⁡ θ + sin^2⁡ θ & cosθ\, sinθ – cosθ sinθ & -sinθ \\cosθ\, sinθ – cosθ sinθ & cos^2⁡ θ + sin^2 ⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\) Solving further, = \(\begin{vmatrix}1 & 0 & -sinθ \\0 & 1 & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\) Breaking the determinant, we get, = 1(0 + cos^2θ) – sinθ(0 – sinθ) = 1

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