Question

The determinant $\left| \begin{matrix} xp + y & x & y \\ yp + z & y & z \\ 0 & xp + y & yp + z \end{matrix} \right|$= 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.

Answer 1

Answer: (B)

x, y, z are in G.P.

Answer 2

D = $\left| \begin{matrix} px + y & x & y \\ py + z & y & z \\ 0 & px + y & py + z \end{matrix} \right|$= 0 Now R₃ ® R₃ – pR₁ – R₂

px + y & x & y \\ py + z & y & z \\ - (xp^{2} + 2py + z) & 0 & 0 \end{matrix} \right|$$ Ž (y² – xz) (p² x + 2py + z ) = 0 Ž y² = xz and x, y, z are in G.P. So choice (2) is true.