2.9k views

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 | 2.9k 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 $C.$

by Active (2.3k points)
edited
+1
nice edit @arjun sir +1 for the edit.....
+22

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

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

+1

@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.
0
$(p+q+r)^3=p^3+q^3+r^3+3(p+q)(q+r)(r+p)$

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

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
by Active (1.2k points)
$\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

by Boss (35.7k points)

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.

by Active (3.2k points)
edited

p=q=r doesn't match

by Active (5.1k points)
0
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
0

@Lone Wolf

$p,q,r$ 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 $p-q=r-p$.

0

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