Question

Let $a,b,c\in R$ be such that $a+b+c\ne 0$ . If the system of equations
$ax+by+cz=0$
$bx+cy+az=0$
$cx+ay+bz=0$
has a non- trivial solution then

• A. $a+c-b=0$
• B. $b+c-a=0$
• C. $a+b-c=0$
• D. $a=b=c$

Answer 1

The correct option is D $a=b=c$

As the system of equations has a non-trivial solution,
$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

Using $C_1\to C_1+C_2+C_3$ we get

$\Delta =(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$

$\Delta =(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& c-b& a-c\\ 0& a-c& b-a\end{array}\right|$

$[Use{C}_3\to C_3-C_2{C}_2\to C_2-C_1]$

$=(a+b+c)[-(b-c\left)\right(b-a)-(a-c{)}^2]$

$=-(a+b+c)[a^2+b^2+c^2-bc-ca-ab]$

$=-\frac{1}{2}(a+b+c)\left[\right(b-c{)}^2+(c-a{)}^2+(a-b{)}^2]$

Now, $\Delta =0,a+b+c\ne 0$

$\Rightarrow {(b-c)}^2+{(c-a)}^2+{(a-b)}^2=0$

As $a,b,c\in R$ We get

$\Rightarrow {(b-c)}^2={(c-a)}^2={(a-b)}^2=0$

$\Rightarrow b-c=0,c-a=0,a-b=0$

$\Rightarrow a=b=c$