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+17 votes

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 (101k points)
edited by | 1.9k views

Another way of solving this question ....

Lets assume option A) is the answer ...

    (i.e)   If the given matrix has non-trivial solution,then A) should be TRUE 

       Taking contra-positive 

             If A) is FALSE,then the given matrix shouldnot have non trivial solution ...

            Let me take p=1,q=1,r=1 (NOTE : Here both conditions in Option A) are false) ... Now determinant value of the given matrix = 0 which means this matrix has non-trivial solution ... So our assumption that "if given matrix has non-trivial solution,then A) should be TRUE" is FALSE ...

The above example not only eliminates A) but also B) and D)

So C) should be the answer ... 

3 Answers

+25 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 (2.3k points)
edited by
nice edit @arjun sir +1 for the edit.....

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


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.

In options

p+q+r=0 matching with answer

but p=q=r never matching


The solution given by @overtomanu is valid ?

not necessary it will satisfy in all situations na?

what if x , y and z has some values such that p , q ,r doesnt necessarily have to be 0 in order to get ouput as 0?



how did this second step come


It should satisfy both equation

but as or is given, so selecting C)

other option is not valid at all





can u explain this one or if you understood this question with some other approach then can you help with that


How did u come conclude that p = q = r is matching for none of the options? For (C) both the conditions in or are matching.
+17 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)
+1 vote




p=q=r doesn't match

answered by Junior (689 points)
why does p = q = r not match ? put p=q=r = k(some constant) in your  $2^{nd}$ equation and verify.

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