Question

Consider the system of equations $2x+P^{2}y+6z=8$ , $x+2y+2qz=5$ and $x+y+3z=4$

• A. Given system has unique solution for $P\ne \pm \sqrt{2}$ and $q=\frac{3}{2}$
• B. Given system has no solution for $P=\pm \sqrt{2}$ and $q=\frac{3}{2}$
• C. Given system has infinite solution for $P=\pm \sqrt{2}$ and $q\in R$
• D. None of these

Answer 1

Answer: (C)

$D=\left|\begin{array}{ccc}2& p^2& 6\\ 1& 2& 2q\\ 1& 1& 3\end{array}\right|=\left(p^2-2\right)(2q-3)$
$D_2=0$
$D_1=\left(p^2-2\right)(8q-15),D_3=\left(P^2-2\right)$