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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. $p - q + r = 0 \text{ or } p = -q = -r$
edited | 1.8k views
+1

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 ...

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
+1
nice edit @arjun sir +1 for the edit.....
+13

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

+2
In options

p+q+r=0 matching with answer

but p=q=r never matching

right?
+1

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?

=3prq−p3−q3−r3

=−(p+q+r)3+3(p+q+r)(pq+qr+pr)

how did this second step come

+1

It should satisfy both equation

but as or is given, so selecting C)

other option is not valid at all

0

@srestha

=3prq−p3−q3−r3

=−(p+q+r)3+3(p+q+r)(pq+qr+pr)

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

+1
@srestha

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.

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)