Question

The system of equations
$2x+py+6z=8$
$x+2y+qz=5$
$x+y+3z=4$
has a unique solution, the value of (p, q) can't be ?

• A. $(1,2)$
• B. $(2,3)$
• C. $(2,4)$
• D. $(3,3)$

Answer 1

The correct option is B $(2,3)$

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 $R_1\to R_1-2R_3$
$=\left|\begin{array}{ccc}0& p-2& 0\\ 1& 2& q\\ 1& 1& 3\end{array}\right|$
$=-(p-2)\left|\begin{array}{cc}1& q\\ 1& 3\end{array}\right|$
$=-(p-2)(3-q)$

$=(2-p)(q-3)$
For unique solution $D\ne 0$ $\Rightarrow p\ne 2$ and $q\ne 3$