Question

The following system of equations
$2x+6y+11=0$,

$6x+20y-6z+3=0$,

$6y-18z+1=0$, is

• A. consistent with unique solution
• B. consistent with infinitely many solution
• C. inconsistent
• D. data insufficient to give the answer

Answer 1

The correct option is C inconsistent

Given:

$2x+6y+11=0$

$6x+20y-6z+3=0$

$6y-18z+1=0$

Converting above equations in the determinant form, we get

$\Delta =\left|\begin{array}{ccc}2& 6& 0\\ 6& 20& -6\\ 0& 6& -18\end{array}\right|$
Here, $\Delta =-648+648=0$
Now,

${\Delta }_1=\left|\begin{array}{ccc}-11& 6& 0\\ -3& 20& -6\\ -1& 6& -18\end{array}\right|$
${\Delta }_1=3276\ne 0$
${\Delta }_2=\left|\begin{array}{ccc}2& -11& 0\\ 6& -3& 6\\ 0& -1& -18\end{array}\right|$
${\Delta }_2=-1092\ne 0$
And ${\Delta }_3=\left|\begin{array}{ccc}2& 6& -11\\ 6& 20& -3\\ 0& 6& -1\end{array}\right|$
${\Delta }_3=-364\ne 0$

So, here $D=0$ and no-one among $D_1,D_2,D_3$ is $0.$

Hence, the system is inconsistent

Hence, option C.