Question

The system of equations
$px+y+z=px+py+z=px+y+pz=p$
has no solution for

• A. three distinct values of $p$
• B. two distinct values of $p$
• C. only one value of $p$
• D. no such $p$ exists

Answer 1

The correct option is C only one value of $p$

$\Delta =\left|\begin{array}{ccc}p& 1& 1\\ 1& p& 1\\ 1& 1& p\end{array}\right|$

$C_2\to C_2-C_3$
$\Rightarrow \Delta =\left|\begin{array}{ccc}p& 0& 1\\ 1& p-1& 1\\ 1& 1-p& p\end{array}\right|$

$\Rightarrow \Delta =p\left[p\right(p-1)-(1-p\left)\right]+1(2-2p)\Rightarrow \Delta =p(p^2-1)-2(p-1)$
$\Rightarrow \Delta =(p-1)(p^2+p-2)$
$\Rightarrow \Delta =(p-1{)}^2(p+2)$

The system has no solution if $\Delta =0$ and at least one of ${\Delta }_1,{\Delta }_2,{\Delta }_3$ is non-zero.

$\Delta =0\Rightarrow p=1,-2$
But, for $p=1$, the system of equations becomes
$x+y+z=1x+y+z=1x+y+z=1$
which has infinite number of solutions.

For $p=-2,{\Delta }_1=-18\ne 0$
So, the system of equations has no solution only for $p=-2$.