If 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 options is TRUE?

Non-Trivial Solution Conditions for Linear System

We are given a system of linear equations:

- px + qy + rz = 0
- qx + ry + pz = 0
- rx + py + qz = 0

This is a homogeneous system of linear equations of the form AX = 0. A homogeneous system has a non-trivial solution (a solution where x, y, and z are not all zero) if and only if the determinant of the coefficient matrix A is equal to zero.

Coefficient Matrix and Determinant Calculation

The coefficient matrix A is:The determinant of A, denoted as det(A), is calculated as follows:

det(A) = p det([[r, p], [p, q]]) - q det([[q, p], [r, q]]) + r * det([[q, r], [r, p]])

det(A) = p(rq - p \cdot p) - q(q \cdot q - p \cdot r) + r(q \cdot p - r \cdot r)

det(A) = p(rq - p^2) - q(q^2 - pr) + r(pq - r^2)

det(A) = pqr - p^3 - q^3 + pqr + pqr - r^3

det(A) = 3pqr - p^3 - q^3 - r^3

det(A) = -(p^3 + q^3 + r^3 - 3pqr)

Condition for Non-Trivial Solution

For the system to have a non-trivial solution, we must have det(A) = 0.

-(p^3 + q^3 + r^3 - 3pqr) = 0

p^3 + q^3 + r^3 - 3pqr = 0

We use the algebraic identity:

p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp)

Therefore, the condition becomes:

(p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp) = 0

This equation holds true if either of the factors is zero:

1. p + q + r = 0
2. p^2 + q^2 + r^2 - pq - qr - rp = 0

Let's analyze the second condition:

p^2 + q^2 + r^2 - pq - qr - rp = 0

Multiply the equation by 2:

2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp = 0

Rearrange the terms to form squares:

(p^2 - 2pq + q^2) + (q^2 - 2qr + r^2) + (r^2 - 2rp + p^2) = 0

(p - q)^2 + (q - r)^2 + (r - p)^2 = 0

Since the square of a real number is always non-negative, the sum of squares can only be zero if each term is individually zero:

- (p - q)^2 = 0 implies p - q = 0, so p = q
- (q - r)^2 = 0 implies q - r = 0, so q = r
- (r - p)^2 = 0 implies r - p = 0, so r = p

Thus, the second condition implies p = q = r.

Conclusion and Option Matching

Combining both conditions, the system has a non-trivial solution if:

p + q + r = 0 OR p = q = r

Comparing this result with the given options, we find that Option 3 matches our derived condition.