First slide

Introduction to Determinants

Question

If $|\begin{array}{ccc} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ \cos^{3} x & \cos^{3} y & \cos^{3} z \end{array}| = 0$ , then which of the following is/are possible?

Moderate

A

x = y
B

y = z
C

x = z
D

x + y + z = $π$ /2

Solution

$|\begin{array}{ccc} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ \cos^{3} x & \cos^{3} y & \cos^{3} z \end{array}|$
Taking cos x, cos y and cos z common from C₁, C₂ and C₃, respectively
$Δ = \cos xcos ycos z |\begin{array}{ccc} \tan x & \tan y & \tan z \\ 1 & 1 & 1 \\ \cos^{2} x & \cos^{2} y & \cos^{2} z \end{array}|$
Applying $C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}$ , we get
$\begin{matrix} Δ = \cos xcos ycos z |\begin{array}{ccc} \tan x - \tan y & \tan y - \tan z & \tan z \\ 0 & 0 & 1 \\ \cos^{2} x - \cos^{2} y & \cos^{2} y - \cos^{2} z & \cos^{2} z \end{array}| \\ = - \cos xcos ycos z |\begin{array}{cc} \tan x - \tan y & \tan y - \tan z \\ \cos^{2} x - \cos^{2} y & \cos^{2} y - \cos^{2} z \end{array}| \\ = - \cos xcos ycos z \times |\begin{array}{cc} \frac{\sin (x - y)}{\cos xcos y} & \frac{\sin (y - z)}{\cos ycos z} \\ - \sin (x - y) \sin (x + y) & - \sin (y - z) \sin (y + z) \end{array}| \\ = \sin (x - y) \sin (y - z) |\begin{array}{cc} \cos z & \cos x \\ \sin (x + y) & \sin (y + z) \end{array}| \\ = \sin (x - y) \sin (y - z) [\cos zsin (y + z) - \cos xsin (x + y)] \\ = \sin (x - y) \sin (y - z) [\sin {ycos}^{2} z + \sin zcos ycos z - \sin xcos ycos x - \sin {ycos}^{2} x] \\ = \sin (x - y) \sin (y - z) [\sin y (\cos^{2} z - \cos^{2} x) + \cos y (\sin zcos z - \sin xcos x)] \\ = \sin (x - y) \sin (y - z) [\sin ysin (x + z) \sin (x - z) + \cos y (\sin 2 z - \sin 2 x) / 2] \\ = \sin (x - y) \sin (y - z) [\sin ysin (x + z) \sin (x - z) + \cos ysin (z - x) \cos (z + x)] \\ = \sin (x - y) \sin (y - z) \sin (z - x) [- \sin ysin (x + z) + \cos ycos (z + x)] \\ = \sin (x - y) \sin (y - z) \sin (z - x) \cos (x + y + z) \end{matrix}$
For $Δ = 0, x = y or y = z or z = x or x + y + z = π / 2$ .

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