GATE CSE 2015 Set 3 | Question: 33

Question

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?

• A. $p - q + r = 0 \text{ or } p = q = -r$
• B. $p + q - r = 0 \text{ or } p = -q = r$
• C. $p + q + r = 0 \text{ or } p = q = r$
• D. $p - q + r = 0 \text{ or } p = -q = -r$

Comments

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 ... 

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If you directly calculate the determinant of the given 3x3 matrix (with no row/col operations), you get the expression $p^3 + q^3 + r^3 – 3pqr = 0$ which is only possible when $p + q + r = 0$, as you can factorize the expression into $(p + q + r)(p^2 + q^2 + r^2 – pq – qr – pr) = 0$

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When for example  p=q=-r,

 p p -p | 0

p -p p | 0

-p p p | 0

by doing row transformations

p p -p | 0

0 0 2p | 0

0 2p 0 | 0

We get, 2py = 2pz = 0

=> y=z=x=0 which gives trivial solutions.

Same happens for option b,d

 

But when p=q=r

After row transformations

p p p | 0

0 0 0 | 0

0 0 0 | 0

Well have to take y = c1, z = c2

c1,c2 are some constants

x = - ( c1 + c2)

Which doesn't have a trivial solution.

Answer 1

 

 

 

p=q=r doesn't match

Comments

for second case

 

(q-r)(p-q) - (r-p)(r-p) = 0

 

 => (q-r)(p-q) = (r-p)(r-p)

 

means

 

1) => q-r = r-p

 

=> 2r = p+q  --------- eq1

 

2) => p-q = r-p

 

=> 2p = r+q

 

or p = (r+q)/2 ---------- eq2

 

put  eq2 in eq1

 

2r = (r+q)/2 + q

 

or 2r = (r + q +2q)/2

 

or 4r = r+3q

 

or 3r = 3q

 

r = q ------------- eq3

 

put eq3 in eq1

 

we get p = 2r/2

 

or p = r ----------------------- eq4

 

using eq3 and eq4

 

r = p = q

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@Lone Wolf

Your approach is wrong.
$p,q,r$ are NOT necessarily integers but they are real numbers i.e. $p,q,r \in \mathbb{R}$.

You can't write from $(q-r)(p-q) = (r-p)(r-p)$ that

$q-r=r-p$ and $p-q=r-p$.

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@aditi19
 

It's indeed $p=q=r$. Check my solution here.