Question-26
If the following system has a non-trivial solution: \[ \begin{aligned} px + qy + rz &= 0\\ qx + ry + pz &= 0\\ rx + py + qz &= 0 \end{aligned} \] Then which one of the following options is true?
This is a homogenous system of linear equations. For this system to have a non-trivial solution, its nullspace should be non-trivial. Now for some row operations: \[ \begin{bmatrix} p & q & r\\ q & r & p\\ r & p & q \end{bmatrix} \xrightarrow[]{R_1 \rightarrow R_1 + R_2 + R_3} \begin{bmatrix} p + q + r & p + q + r & p + q + r\\ q & r & p\\ r & p & q \end{bmatrix} \] If \(p + q + r = 0\) then the nullity is at least \(1\). So this is one condition. Let us now assume that \(p + q + r \neq 0\) and proceed: \[ \begin{bmatrix} p + q + r & p + q + r & p + q + r\\ q & r & p\\ r & p & q \end{bmatrix} \xrightarrow[]{R_1 \rightarrow \frac{1}{p + q + r} R_1} \begin{bmatrix} 1 & 1 & 1\\ q & r & p\\ r & p & q \end{bmatrix} \] We can now proceed further: \[ \begin{bmatrix} 1 & 1 & 1\\ q & r & p\\ r & p & q \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1\\ 0 & r - q & p - r\\ 0 & p - r & q - r \end{bmatrix} \] The nullity can be non-zero if one of the last two rows is zero or one of them is a multiple of the other. If the second or third row is zero, we get \(p = q = r\). If not, then we have: \[ r - q = k(p - r) \text{ and } p - r = k(q - r) \] This is impossible for any real \(k\). Therefore, we have: \[ \boxed{p + q + r = 0 \text{ or } p = q = r} \]