If the system of linear equations
$x+2ay + az = 0$
$x + 3by + bz = 0$
$x + 4cy + cz = 0$
has a non-zero solution, then a, b, c
Answer & explanation
Correct answer: option 4
The given system of equations will have a non-zero solution, if
$\begin{vmatrix}1 &2a &a\\1 &3b& b\\1 &4c&c\end{vmatrix}=0$
$⇒\begin{vmatrix}1 &2a &a\\0 &3b-2a& b-a\\0 &4c-2a&c-a\end{vmatrix}=0$ Applying $R_2 → R_2-R_1, R_3 → R_3-R_1$
$⇒ (3b-2a) (c-a) - (b-a) (4c - 2a) = 0$
$⇒ (3bc - 3ab - 2ca + 2a^2) - (4bc - 2ab - 4ca + 2a^2) = 0$
$⇒-bc - ab + 2ca = 0$
$⇒\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ ⇒ a, b, c are in H.P.