Question

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

• A. x, y and z are in GP
• B. x, y and z are in AP
• C. x, y and z are in HP
• D. xy, yz, zx are in AP

Answer 1

Answer: (A)

x, y and z are in GP

Answer 2

The given determinant is $D = \begin{vmatrix} xp + y & x & y \\ yp + z & y & z \\ 0 & xp + y & yp + z \end{vmatrix}$ Expanding the determinant along the third row: $D = 0 \cdot C_{31} - (xp+y) \cdot C_{32} + (yp+z) \cdot C_{33}$ where $C_{ij}$ is the cofactor of the element in the $i$-th row and $j$-th column. $C_{31} = \begin{vmatrix} x & y \\ y & z \end{vmatrix} = xz - y^2$ $C_{32} = \begin{vmatrix} xp+y & y \\ yp+z & z \end{vmatrix} = (xp+y)z - y(yp+z) = xpz + yz - y^2p - yz = xpz - y^2p = p(xz - y^2)$ $C_{33} = \begin{vmatrix} xp+y & x \\ yp+z & y \end{vmatrix} = (xp+y)y - x(yp+z) = xpy + y^2 - xyp - xz = y^2 - xz = -(xz - y^2)$ Substituting these cofactors into the determinant expansion: $D = 0 \cdot (xz - y^2) - (xp+y) \cdot p(xz - y^2) + (yp+z) \cdot (-(xz - y^2))$ $D = - p(xp+y)(xz - y^2) - (yp+z)(xz - y^2)$ Factoring out $(xz - y^2)$: $D = -(xz - y^2) [p(xp+y) + (yp+z)]$ $D = -(xz - y^2) [xp^2 + yp + yp + z]$ $D = -(xz - y^2) [xp^2 + 2yp + z]$ The determinant is equal to zero, i.e., $D = 0$. $-(xz - y^2) [xp^2 + 2yp + z] = 0$ This equation holds if either $(xz - y^2) = 0$ or $[xp^2 + 2yp + z] = 0$.

The first condition, $xz - y^2 = 0$, means $y^2 = xz$. This is the condition for $x, y, z$ to be in Geometric Progression (GP). If $x, y, z$ are in GP, then $y^2 = xz$, which makes the first factor zero, and thus the determinant is zero for any value of $p$.

The second condition, $xp^2 + 2yp + z = 0$, is a quadratic equation in $p$. If this equation holds for some value of $p$, the determinant is zero. However, the options provided are relationships between $x, y, z$ that are independent of $p$. The question asks for a condition on $x, y, z$ for which the determinant is zero. This suggests that the determinant should be zero based on a relationship between $x, y, z$ that holds for any value of $p$ (for which the expressions are defined). This happens when the factor $(xz - y^2)$ is zero.

If $x, y, z$ are in GP, then $y^2 = xz$, so $xz - y^2 = 0$, which makes the determinant zero regardless of the value of $p$.

Therefore, x, y, and z are in GP.