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

non-trivial solution so determinant = 0, 

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it may helps you 

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Concept:
Homogeneous system mein unique solution=ONLY X=0.
Non-trivial solution exists⟺det(A)=0⟺Free variable exists.​
AX=0 has non-trivial solution⟺det(A)=0⟺rank(A)<n⟺∞ solutions.​

Answer 1

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 $C.$

Comments

@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.

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$(p+q+r)^3=p^3+q^3+r^3+3(p+q)(q+r)(r+p)$

$=p^3+q^3+r^3+3(p+q+r)(pq+qr+rp)-3pqr$

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a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−ac−bc)
from this it is clear that if (a+b+c)=0,then a3+b3+c3−3abc=0 and a3+b3+c3=3abc.

Also if a=b=c , a3+b3+c3=a3+a3+a3=3a3=3aaa=3abc,

hence a3+b3+c3=3abc
 

Answer 2

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

Answer 3

Let $A=\begin{bmatrix}p & q & r\\q & r & p\\r & p & q \end{bmatrix}$.

Since the system has non-trivial solution, hence

$\begin{align} |A|&=0\\ \Rightarrow \left|\begin{matrix}p & q & r\\q & r & p\\r & p & q \end{matrix}\right| &=0\\ \Rightarrow \left|\begin{matrix}p+q+r & q & r\\p+q+r & r & p\\p+q+r & p & q \end{matrix}\right| &=0;~[C_1 \leftarrow  C_1+C_2+C_3]\\ \Rightarrow (p+q+r)\left|\begin{matrix}1 & q & r\\1 & r & p\\1 & p & q \end{matrix}\right| &=0\\ \Rightarrow (p+q+r)\left\{ (qr-p^2) -(q^2-pr)+(pq-r^2)  \right\} &=0; ~[\scriptsize\text{Expanding the determinant using the 1st column}]\\ \Rightarrow (p+q+r)(p^2+q^2+r^2-pq-qr-rp) &=0 \end{align}$

 

$\therefore p+q+r=0\mathrm{~Or~}(p^2+q^2+r^2-pq-qr-rp)=0~...............\tag{i}$

 

Now

$\begin{align} p^2+q^2+r^2-pq-qr-rp&=0\\ \Rightarrow 2p^2+2q^2+2r^2-2pq-2qr-2rp &=0;~[\scriptsize\text{Multiplying both sides by }2]\\ \Rightarrow (p^2-2pq+q^2)+(q^2-2qr+r^2)+(r^2-2rp+p^2) &=0;~[\scriptsize\text{Rearranging the expression}]\\ \Rightarrow (p-q)^2+(q-r)^2+(r-p)^2 &=0\\ \therefore p-q=0 \mathrm{~And~}q-r=0 \mathrm{~And~}r-p&=0;~[\scriptsize\text{Since the sum of some squares is }0 \text{, hence they must be separately }0.] \\ \Rightarrow p=q \mathrm{~And~}q=r \mathrm{~And~}r=p\\ \Rightarrow p=q=r \end{align}$

 

Therefore, from no$\mathrm{(i)}$, we can conclude that

$$p+q+r=0 \mathrm{~Or~} p=q=r$$.

 

So the correct answer is C.

Comments

Same as I was hoping to see :)

Answer 4

$\begin{bmatrix} p & q &r \\ q & r & p\\ r & p & q \end{bmatrix}$$\begin{bmatrix} x\\ y\\ z \end{bmatrix} = $ $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

$R_1 \rightarrow R_1 + R_2 + R_3$

$\begin{bmatrix} p+q+r & p+q+r &p+q+r \\ q & r & p\\ r & p & q \end{bmatrix}$

option C is $p + q + r = 0 \text{ or } p = q = r$

$\begin{bmatrix} 0 & 0 & 0 \\ q & r & p\\ r & p & q \end{bmatrix}$$\begin{bmatrix} x\\ y\\ z \end{bmatrix} = $ $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

For non-trivial solution,

if $AX = 0$ then Rank(A) $\leq$ number of unknowns

2 $\leq$ 3

Hence, C is the answer

Answer 5

 

 

 

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.

Answer 6

Ans