Question

If x,y,z are in GP, using properties of determinants, show that

$\left|\begin{array}{ccc}px+y& x& y\\ py+z& y& z\\ 0& px+y& py+z\end{array}\right|=0$, where $x\ne y\ne z$ and p is any real number.

Answer 1

Given,

$x,y,zareinGP⟹y^2=xz$

LHS :

$\left|\begin{array}{ccc}px+y& x& y\\ py+z& y& z\\ 0& px+y& py+z\end{array}\right|$

Applying $C_1\to C_1-p{C}_2-C_3$, we get,

$=\left|\begin{array}{ccc}0& x& y\\ 0& y& z\\ -p^{2}x-py-py-z& px+y& py+z\end{array}\right|$

$=\left|\begin{array}{ccc}0& x& y\\ 0& y& z\\ -p^{2}x-2py-z& px+y& py+z\end{array}\right|$

On expanding along $C_1$, we get,

$=-p^{2}x-2py-z(xz-y^2)$

$=(-p^{2}x-2py-z)\times 0=0=RHS$

Hence, proved.