Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Determinants

The determinant $\begin{vmatrix}xp+y&x&y\\yp+z&y&z\\0 &xp+y&yp+z\end{vmatrix}=0$ if

(A) x,y,z are in A.P

(B) x,y,z are in G.P

(C) x,y,z are in H.P

(D) xy,yz,zx are in A.P

1 Answer

Given :
$\begin{vmatrix}xp+y&x&y\\yp+z&y&z\\0 &xp+y&yp+z\end{vmatrix}=0$
operating $C_1=C_1-PC_2-C_3$
$\begin{vmatrix}0&x&y\\0&y&z\\-(xp^2+2py+z) &xp+y&yp+z\end{vmatrix}=0$
$\Rightarrow (xz-y^2)(xp^2+2py+z)=0$
$\Rightarrow xz-y^2=0$
$\Rightarrow D=0$ for other quad.factor in p which shows second factor is a perfect square.
$\Rightarrow y^2=xz$
$\Rightarrow x,y,z$ are in G.P.
Hence (b) is the correct answer.
answered Nov 20, 2013 by sreemathi.v
edited Jun 17, 2014 by rohanmaheshwari0831_1

Related questions