Question

The following system has non-trivial solution $ px + qy + rz = 0, qx + ry + pz = 0, rx + py + qz = 0 $ then which one of the following is TRUE?

• A. \( p - q + r = 0 \) or \( p = q = -r \)
• B. \( p + q - r = 0 \) or \( p = -q = r \)
• C. \( p + q + r = 0 \) or \( p = q = r \)
• D. \( p - q - r = 0 \) or \( p = -q = -r \)

Hint:

For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

Answer 1

The Correct Option is C

The given system of equations is a homogeneous system, which means that for a non-trivial solution (i.e., a solution where not all variables are zero), the determinant of the coefficient matrix must be zero. The determinant of the coefficient matrix for the given system is:
\[ \begin{vmatrix} p & q & r
q & r & p
r & p & q \end{vmatrix} = 0 \] The solution to this determinant equation leads to the condition that either:

(1) \( p + q + r = 0 \), or

(2) \( p = q = r \).
Thus, the correct condition that must be true for the non-trivial solution is \( p + q + r = 0 \) or \( p = q = r \).

(1) Why Other Options Are Incorrect:
- Option 1: \( p - q + r = 0 \) is incorrect because it does not correspond to the determinant condition.
- Option 2: \( p + q - r = 0 \) does not satisfy the determinant equation.
- Option 4: \( p - q - r = 0 \) is also incorrect.

Conclusion: The correct condition for the given system of equations to have a non-trivial solution is \( p + q + r = 0 \) or \( p = q = r \).