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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?

 

  1. $p - q + r = 0 \text{ or } p = q = -r$
  2. $p + q - r = 0 \text{ or } p = -q = r$
  3. $p + q + r = 0 \text{ or } p = q = r$
  4. $p - q + r = 0 \text{ or } p = -q = -r$
asked in Linear Algebra by Veteran (76.2k points)   | 808 views

2 Answers

+12 votes
Best answer

for non-trivial solution $$\left | A \right | = 0$$

where $\left | A \right | = \begin{bmatrix} p & q& r\\ q& r& p\\ r& p & q \end{bmatrix} = p*(rq-p^{2})-q*(q^{2}-pr)+r*(qp-r^{2})$

$=prq - p^3 - q^3 + prq + prq - r^3 \\= 3prq - p^3 - q^3 - r^3 \\=-{\left(p+q+r\right)}^3 + 3(p+q+r)(pq+qr+pr)$

now if you check the options the only options where each individual condition can make $\left | A \right | = 0$ zero is C

answered by Active (2.2k points)  
selected by
nice edit @arjun sir +1 for the edit.....

for a homogeneous equation to have a consistent solution, the general equation:

$AX=0$

must be satisfied.
If $A^{-1}$ exists then we can multiply by $A^{-1}$ on both sides and get $X=0$, which means solution is trivial.

But if $A^{-1}$ does not exist, meaning $|A|=0$, we cannot multiply both sides by $A^{-1}$ to reach $X=0$. In which case it implies that other non-trivial solutions exists.

http://math.stackexchange.com/questions/1012571/non-trivial-solutions-for-homogeneous-equations

+8 votes
Answer = C

If we add all the equations we get

(p+q+r)x + (p+q+r)y + (p+q+r)z = 0

which implies p+q+r=0

Only option C has p+q+r=0
answered by Active (1.2k points)  
Answer:

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