Q9 of 26 Page 1

If x, y, z are in GP, then using properties of determinants, show that , where x ≠ y ≠ z and p is any real number.

Given: x, y, z are in G.P


To show:


Proof:



C1 C1 – pC2 – C3





Expanding by R3


Δ = (-p2y – 2py – z)(xz – y2) …(i)


Since, x, y, z are in G.P. This means



y2 = xz


y2 – xz = 0


or xz – y2 = 0


Substituting the value in eq. (i), we get


Δ = (-p2y – 2py – z)(0)


Δ = 0


LHS = RHS


Hence Proved


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