The determinant , if
- A.are in AP
- B.are in GP
- C.are in HP
- D.are in AP
Show answer and solution
Correct option: B
Solution
Expand the determinant along the third row or use cofactor expansion. The determinant equals:
0·(minor) - (xp+y)·[(xp+y)z - y(yp+z)] + (yp+z)·[(xp+y)y - x(yp+z)]
Compute each bracket:
- (xp+y)z - y(yp+z) = xpz + yz - y²p - yz = p(xz - y²)
- (xp+y)y - x(yp+z) = xy·p + y² - xy·p - xz = y² - xz
So the determinant = -(xp+y)·p(xz - y²) + (yp+z)·(y² - xz)
= (y² - xz)[p(xp+y) - (yp+z)]... wait, factor out (y² - xz):
= (y² - xz)[-(xp+y)p + (yp+z)]
= (y² - xz)[-xp² - yp + yp + z]
= (y² - xz)(z - xp²)
For this to be zero for all p (or as a condition independent of p), we need y² - xz = 0, i.e., y² = xz, which means x, y, z are in GP.
0·(minor) - (xp+y)·[(xp+y)z - y(yp+z)] + (yp+z)·[(xp+y)y - x(yp+z)]
Compute each bracket:
- (xp+y)z - y(yp+z) = xpz + yz - y²p - yz = p(xz - y²)
- (xp+y)y - x(yp+z) = xy·p + y² - xy·p - xz = y² - xz
So the determinant = -(xp+y)·p(xz - y²) + (yp+z)·(y² - xz)
= (y² - xz)[p(xp+y) - (yp+z)]... wait, factor out (y² - xz):
= (y² - xz)[-(xp+y)p + (yp+z)]
= (y² - xz)[-xp² - yp + yp + z]
= (y² - xz)(z - xp²)
For this to be zero for all p (or as a condition independent of p), we need y² - xz = 0, i.e., y² = xz, which means x, y, z are in GP.
[1997]