The determinant ∣∣
∣∣xp+yxyyp+zyz0xp+yyp+z∣∣
∣∣=0 if
A
x,y,z are in A.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x,y,z are in G.P.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x,y,z are in H.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
xy,yz,zx are in A.P.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Bx,y,z are in G.P. Given that, ∣∣
∣∣xp+yxyyp+zyz0xp+yyp+z∣∣
∣∣=0 Operating C1→C1−pC2−C3, we get ∣∣
∣
∣∣0xy0yz−(xp2+2py+z)xp+yyp+z∣∣
∣
∣∣=0or,(xz−y2)(xp2+2py+z)=0or,xz−y2=0or,y2=xzx,y,z are in G.P.