Question

If \( \alpha + \beta + \gamma = 2\pi \), then the system of equations
\( x + (\cos \gamma)y + (\cos \beta)z = 0 \)
\( (\cos \gamma)x + y + (\cos \alpha)z = 0 \)
\( (\cos \beta)x + (\cos \alpha)y + z = 0 \)
has :

• A. infinitely many solutions
• B. no solution
• C. a unique solution
• D. exactly two solutions

Hint:

The condition \( \cos^2 A + \cos^2 B + \cos^2 C - 2\cos A \cos B \cos C = 1 \) holds whenever \( A+B+C = 2\pi \) or \( A+B+C = \pi \).
In such cases, the homogeneous system formed by these cosines always has non-trivial solutions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a homogeneous system of linear equations of the form \( AX = 0 \).
Such a system has a non-trivial solution (infinitely many solutions) if and only if the determinant of the coefficient matrix \( |A| = 0 \).
Step 2: Key Formula or Approach:
Calculate the determinant \( D \):
\[ D = \begin{vmatrix} 1 & \cos \gamma & \cos \beta
\cos \gamma & 1 & \cos \alpha
\cos \beta & \cos \alpha & 1 \end{vmatrix} \]
Step 3: Detailed Explanation:
Expanding the determinant:
\[ D = 1(1 - \cos^2 \alpha) - \cos \gamma(\cos \gamma - \cos \alpha \cos \beta) + \cos \beta(\cos \alpha \cos \gamma - \cos \beta) \]
\[ D = \sin^2 \alpha - \cos^2 \gamma + \cos \alpha \cos \beta \cos \gamma + \cos \alpha \cos \beta \cos \gamma - \cos^2 \beta \]
\[ D = 1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma \]
For \( \alpha + \beta + \gamma = 2\pi \), we use the identity \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma - 2\cos \alpha \cos \beta \cos \gamma = 1 \).
Substituting this identity:
\[ D = 1 - 1 = 0 \]
Since \( D = 0 \), the system has infinitely many solutions.
Step 4: Final Answer:
The system has infinitely many solutions.