Question

If the system of linear equations
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$
has infinitely many solutions, then a possible pair of $p$ and $q$ is

• A. $p=3,q=3$
• B. $p\ne 2,q\ne 3$
• C. $p\ne 2,q=3$
• D. $p=2,q\ne 3$

Answer 1

The correct option is D $p=2,q\ne 3$

$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$

For infinitely many solutions,
$\Delta ={\Delta }_x={\Delta }_y={\Delta }_z=0$

$\Delta =\left|\begin{array}{ccc}2& p& 6\\ 1& 2& q\\ 1& 1& 3\end{array}\right|=(p-2)(q-3)$

${\Delta }_x=\left|\begin{array}{ccc}8& p& 6\\ 5& 2& q\\ 4& 1& 3\end{array}\right|$

$=30-8q-15p+4pq=(p-2)(4q-15)$

${\Delta }_y=\left|\begin{array}{ccc}2& 8& 6\\ 1& 5& q\\ 1& 4& 3\end{array}\right|=0$

${\Delta }_z=\left|\begin{array}{ccc}2& p& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\right|=p-2$

At $p=2,\Delta ={\Delta }_x={\Delta }_y={\Delta }_z=0$
So, at $p=2$ and $q\in R$, the system of linear equations has infinitely many solutions.