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