Question

For what values of $p$ and $q$, the system of equations $2x+py+6z=8,x+2y+qz=5,x+y+3z=4$ has unique solution.

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

Answer 1

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

Given equations are
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$
Here $D=\left|\begin{array}{ccc}2& p& 6\\ 1& 2& q\\ 1& 1& 3\end{array}\right|$
Applying $R_1\to R_1-2R_3$
$=\left|\begin{array}{ccc}0& p-2& 0\\ 1& 2& q\\ 1& 1& 3\end{array}\right|$
$=-(p-2)\left|\begin{array}{cc}1& q\\ 1& 3\end{array}\right|$
$=-(p-2)(3-q)$ $=(p-2)(q-3)$
For unique solution $D\ne 0$ $\Rightarrow p\ne 2$ and $q\ne 3$