Question

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) 2

The question was asked in an interview.

Question is taken from Determinant topic in chapter Determinants of Mathematics – Class 12

Answer 1

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