Question

If the system of equations,
$x-2y+z=a$
$2x+y-2z=b$
$x+3y-3z=c$
have atleast one solution, then the relationship between $a,b$ and $c$ is

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

Answer 1

The correct option is B $a-b+c=0$

Given:

$x-2y+z=a$

$2x+y-2z=b$

$x+3y-3z=c$

$\Delta =\left|\begin{array}{ccc}1& -2& 1\\ 2& 1& -2\\ 1& 3& -3\end{array}\right|$

$=1(-3+6)+2(-6+2)+1(6-1)$

$=3-8+5=0$

As $\Delta =0$, system does not have a unique solution.

To have at least one solution, $\Delta r=\Delta y=\Delta z=0$ (infinit solutions)

$\Delta x=\left|\begin{array}{ccc}a& -2& 1\\ b& 1& -2\\ c& 3& -3\end{array}\right|$

$a(-3+6)+2(-3b+2c)+(3b-c)=0$

$\Rightarrow +3a-6b+3b+4c-c=0$

$a-b+c=0$

$\therefore$ option B is incorrect