Question

For the system of equations given by, $2x+py+6z=8,x+2y+qz=5,x+y+3z=4$, which of the following is/are true:

• A. For $p\ne 2$,$q\ne 3$, the system has unique solution.
• B. For $p=2,q\in R,$ the system has infinitely many solutions.
• C. For $p\ne 2,q=3,$ the system has no solution.
• D. Can't be determined.

Answer 1

Answer: (A), (B), (C)

For $p\ne 2,q=3,$ the system has no solution.

The given system of equation is
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$

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

By Cramer's rule, if $\Delta \ne 0$, i.e., $p\ne 2$,$q\ne 3$, the system has unique solution.

If $p=2orq=3,\Delta =0$, then if ${\Delta }_x={\Delta }_y={\Delta }_z=0$, the system has infinite solutions and if any one of ${\Delta }_x,{\Delta }_y,{\Delta }_z\ne 0$, system has no solution. Now,

${\Delta }_x=\left|\begin{array}{ccc}8& p& 6\\ 5& 2& q\\ 4& 1& 3\end{array}\right|=30-8q-15p+4pq=(p-2)(4q-15)$

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

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

Thus if $p=2,{\Delta }_x={\Delta }_y={\Delta }_z=0$ for all $q\in R$, So the system has infinite solutions.

And if $p\ne 2,q=3,{\Delta }_x,{\Delta }_z\ne 0$, the system has no solution.

Hence the system has,
(i) no solution, if $p\ne 2,q=3,$
(ii) a unique solution, if $p\ne 2,q\ne 3,$
(iii) infinitely many solutions, for $p=2,q\in R$