Question

If the system of linear equations
$2x+2ay+az=02x+3by+bz=02x+4cy+cz=0,$

where $a,b,c\in R$ are non-zero and distinct; has non-zero solution, then

• A. $a+b+c=0$
• B. $a,b,c$ are in A.P.
• C. $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
• D. $a,b,c$ are in G.P.

Answer 1

The correct option is C $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

$\left|\begin{array}{ccc}2& 2a& a\\ 2& 3b& b\\ 2& 4c& c\end{array}\right|=0$

$R_2\to R_2-R_1,R_3-R_1\Rightarrow \left|\begin{array}{ccc}2& 2a& a\\ 0& 3b-2a& b-a\\ 0& 4c-2a& c-a\end{array}\right|=0$

$\Rightarrow 3bc-2ac-3ab+2a^2-[4bc-4ac-2ab+2a^2]=0\Rightarrow -bc+2ac-ab=0\Rightarrow ab+bc=2ac\Rightarrow \frac{1}{c}+\frac{1}{a}=\frac{2}{b}$
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.