Question

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

Answer 1

$2x+py+6z=8$

$x+2y+qz=5$

$x+y+3z=4$

$\Rightarrow △\ne 0$

Unique solution$△=\left|\begin{array}{ccc}2& p& 6\\ 1& 2& q\\ 1& 1& 3\end{array}\right|\ne 0$

$\Rightarrow 12-2q-3p+pq-6\ne 0$

$=(p-2)(q-3)\ne 0$

$\Rightarrow p\ne 2q\ne 3$