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34 votes
34 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. $p - q + r = 0 \text{ or } p = -q = -r$
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5 Answers

Best answer
41 votes
41 votes

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 $C.$

edited by
43 votes
43 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
23 votes
23 votes

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.

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

$R_1 \rightarrow R_1 + R_2 + R_3$

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

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

$\begin{bmatrix} 0 & 0 & 0 \\ p & 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:

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