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Worked Examples · Example 22

Q.Evaluate Δ=∣0cos⁡x−sin⁡x−cos⁡x0cos⁡ysin⁡x−cos⁡y0∣\Delta = \begin{vmatrix} 0 & \cos x & -\sin x \\ -\cos x & 0 & \cos y \\ \sin x & -\cos y & 0 \end{vmatrix}.

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✓ Free question

Expanding along the first row, the two non-zero terms cancel exactly, so the determinant is 00 (as expected for an odd-order skew-symmetric determinant).

Δ=a11A11+a12A12+a13A13,Aij=(−1)i+jMij.\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13},\qquad A_{ij} = (-1)^{i+j}M_{ij}.

  1. Expand along Row 1 of Δ=∣0cos⁡x−sin⁡x−cos⁡x0cos⁡ysin⁡x−cos⁡y0∣\Delta = \begin{vmatrix} 0 & \cos x & -\sin x \\ -\cos x & 0 & \cos y \\ \sin x & -\cos y & 0 \end{vmatrix}. Since a11=0a_{11}=0, that term vanishes.
  2. Term with a12=cos⁡xa_{12} = \cos x: minor M12=∣−cos⁡xcos⁡ysin⁡x0∣=(−cos⁡x)(0)−(cos⁡y)(sin⁡x)=−sin⁡xcos⁡y.M_{12} = \begin{vmatrix} -\cos x & \cos y \\ \sin x & 0 \end{vmatrix} = (-\cos x)(0) - (\cos y)(\sin x) = -\sin x\cos y. Cofactor A12=−M12=sin⁡xcos⁡y.A_{12} = -M_{12} = \sin x\cos y. Contribution =cos⁡x (sin⁡xcos⁡y).= \cos x\,(\sin x\cos y).
  3. Term with a13=−sin⁡xa_{13} = -\sin x: minor M13=∣−cos⁡x0sin⁡x−cos⁡y∣=(−cos⁡x)(−cos⁡y)−(0)(sin⁡x)=cos⁡xcos⁡y.M_{13} = \begin{vmatrix} -\cos x & 0 \\ \sin x & -\cos y \end{vmatrix} = (-\cos x)(-\cos y) - (0)(\sin x) = \cos x\cos y. Cofactor A13=+M13=cos⁡xcos⁡y.A_{13} = +M_{13} = \cos x\cos y. Contribution =−sin⁡x (cos⁡xcos⁡y).= -\sin x\,(\cos x\cos y).
  4. Add the contributions:

Δ=cos⁡xsin⁡xcos⁡y−sin⁡xcos⁡xcos⁡y=0.\Delta = \cos x\sin x\cos y - \sin x\cos x\cos y = 0.

✓Final answer

Δ=0\Delta = 0.

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