Question

The determinant is $ \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 $AP$
• B. $x, y, z$ are in $GP$
• C. $x, y, z$ are in $HP$
• D. $xy, yz, zx$ are in $AP$

Answer 1

Answer: (B)

We have, $\begin{vmatrix}xp+y&x&y\\ yp+z&y&z\\ 0&xp+y&yp+z\end{vmatrix} = 0 $
$ \Rightarrow \begin{vmatrix}xp&x&y\\ yp&y&z\\ -yp-z&xp+y&yp+z\end{vmatrix} = 0 $
$[$Applying $C_1 \to C_1 - C_3]$
$\Rightarrow \begin{vmatrix}0&x&y\\ 0&y&z\\ -z-xp^{2}&xp+y&yp+z\end{vmatrix} = 0 $
$[$ Applying $C_1 \to C_1 - p\,C_2]$
$\Rightarrow 0 - 0 + (- z - xp^2) [ xz - y^2] = 0$
$[$ Expanding along $C_1]$
$\Rightarrow -(z - xp^2) [ xz - y^2] = 0$
$\Rightarrow z - xp^2 = 0$ or $xz - y^2 = 0$
$\therefore y^2 = xz$
Hence, $ x, y, z $ are in $GP$.