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

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

=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

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

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

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} = $ $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

$R_1 \rightarrow R_1 + R_2 + R_3$

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

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