Question

If the system of linear equations $x+ay+az=0,x+by+bz=0,x+cy+cz=0$ has a non-zero solution then

• A. System has always non-trivial solutions
• B. System is consistent only when $a=b=c$
• C. If $a\ne b\ne c$ then $x=0,y=t,z=-t\forall tϵR$
• D. If $a=b=c$ then $y=t_1,z=t_2,x=-a(t_1+t_2)\forall t_1,t_2ϵR$

Answer 1

Answer: (A), (C), (D)

The correct options are
A System has always non-trivial solutions
C If $a\ne b\ne c$ then $x=0,y=t,z=-t\forall tϵR$
D If $a=b=c$ then $y=t_1,z=t_2,x=-a(t_1+t_2)\forall t_1,t_2ϵR$

Given system of equations has a non-zero solution
$x+ay+az=0$ ....(1)
$x+by+bz=0$ .....(2)
$x+cy+cz=0$ ....(3)
It can be written as $AX=O$
where $A=\left[\begin{array}{ccc}1& a& a\\ 1& b& b\\ 1& c& c\end{array}\right]$
Clearly $|A|=0$
So, the system has infinitely many solutions and hence it has non-trivial solutions.
Since, the system has infinite many solutions.
So, let $y=t;t\in R$
Then eqn (1) and (2),
$x+az=-at$
$x+bz=-bt$
Solving these, we get
$z=-t;x=0$
Hence, if $a\ne b\ne c$ then $x=0;y=t,z=-t;t\in R$
If $a=b=c$
Let $y=t_1,z=t_2;t_1,t_2\in R$
Then by eqn (1),
$x=-a(t_1+t_2)$