Question

Consider the system of equations
$kx+(c-1)y+z=2$
$cx+(k+1)y+kz=4$
$x+cy+z=1$

Then correct statement is/are

• A. If $c=1$, then the system will be inconsistent.
• B. If $c=1$, then the may have infinite solutions.
• C. If $k=2\cos \frac{\pi }{5}$ then the system will be inconsistent.
• D. The system can never possess infinite solutions.

Answer 1

Answer: (B), (C)

The correct options are
B If $c=1$, then the may have infinite solutions.
C If $k=2\cos \frac{\pi }{5}$ then the system will be inconsistent.

$\Delta =\left|\begin{array}{ccc}k& c-1& 1\\ c& k+1& k\\ 1& c& 1\end{array}\right|=k\left(\right(k+1)-ck)-(c-1)(c-k)+(c^2-(k+1\left)\right)$
$=(1-c)k^2+(c-1)k+(c-1)$
$=(1-c)(k^2-k-1)$
If $\Delta =0\Rightarrow c=1ork=\frac{1\pm \sqrt{1+4}}{2}\Rightarrow k=\frac{1\pm \sqrt{5}}{2}k=\frac{1+\sqrt{5}}{2}=2cos\frac{\pi }{5}$
${\Delta }_1=\left|\begin{array}{ccc}2& c-1& 1\\ 4& k+1& k\\ 1& c& 1\end{array}\right|=2(k+1-ck)-(c-1)(4-k)+1(4c-k-1)$
$=2k+2-2ck-4c+ck+4-k+4c-k-1=-ck+5$
For c =1 and infinitely many solutions,
${\Delta }_1=0\Rightarrow (5-k)=0\Rightarrow k=5$

${\Delta }_2=\left|\begin{array}{ccc}k& 2& 1\\ c& 4& k\\ 1& 1& 1\end{array}\right|=k(4-k)-2(c-k)+1(c-4)$
$=4k-k^2-2c+2k+c-4$
$=(-k^2+6k-c-4)$
For c =1 and infinitely many solutions,
${\Delta }_2=0\Rightarrow -(k^2-6k+5)=0\Rightarrow k=1or5$
${\Delta }_3=\left|\begin{array}{ccc}k& c-1& 2\\ c& k+1& 4\\ 1& c& 1\end{array}\right|=k(k+1-4c)-(c-1)(c-4)+2(c^2-k-1)=k^2+k-4ck-c^2+4c+c-4+2c^2-2k-2=k^2+c^2-4ck-k+5c-6$
For c =1 and infinitely many solutions,
${\Delta }_3=0\Rightarrow k^2-5k=0\Rightarrow k=0or5$
For c = 1 and k = 5, we have
${\Delta }_1,{\Delta }_2,{\Delta }_3=0$
Therefore, the given system of equations can have infinitely many solutions for c = 1 and k =5.
$Ifk=2cos\frac{\pi }{5}$
${\Delta }_1,{\Delta }_2,{\Delta }_3$ are not simultaneously zero for any value c, so the system will be inconsistent.