Question

If the system of equations
$cx+y+1=0$
$x+cy+2=0$
$x+y+1=0$
is consistent, then the value of $c$ can be:

• A. $2$
• B. $3$
• C. $1$
• D. $5$

Answer 1

The correct option is A $2$

$D=\left|\begin{array}{ccc}c& 1& 1\\ 1& c& 2\\ 1& 1& 1\end{array}\right|=0$
$C_3\to C_3-C_2$
$\Rightarrow \left|\begin{array}{ccc}c& 1& 0\\ 1& c& 2-c\\ 1& 1& 0\end{array}\right|=0$
$\Rightarrow (2-c)(c-1)=0\Rightarrow c=1,2$
Now, for $c=1$ system becomes
$x+y+1=0$
$x+y+2=0$
$x+y+1=0$
which is inconsistent system (as lines are parallel).

Now, for $c=2$ system becomes
$2x+y+1=0$
$x+2y+2=0$
$x+y+1=0$
which is consistent system with solution $(0,-1).$