If system of equations2x+2ay+az=02x+3by+bz=02x+4cy+cz=0 has a non-trival 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.

Solution:

If the given system of equations have a non-trivial

solution, then

$\begin{array}{l} |\begin{array}{lll} 2 & 2 a & a \\ 2 & 3 b & b \\ 2 & 4 c & c \end{array}| = 0 \Rightarrow 2 |\begin{array}{lll} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{array}| = 0 \\ Applying R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} \\ = 2 |\begin{array}{ccc} 1 & 2 a & a \\ 0 & 3 b - 2 a & b - a \\ 0 & 4 c - 2 a & c - a \end{array}| = 0 \end{array}$

$\begin{array}{l} = 2 |\begin{array}{ll} 3 b - 2 a & b - a \\ 4 c - 2 a & c - a \end{array}| = 0 \\ = (3 b - 2 a) (c - a) - (4 c - 2 a) (b - a) = 0 \\ = (3 b c - 3 a b - 2 a c + 2 a^{2}) - (4 b c - 4 c a - 2 a b + 2 a^{2}) = 0 \\ = - b c - a b + 2 c a = 0 \\ = \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A.P. \end{array}$

If system of equations $\begin{aligned} 2 x + 2 a y + a z = 0 \\ 2 x + 3 b y + b z = 0 \\ 2 x + 4 c y + c z = 0 \end{aligned}$ has a non-trival solution, then

Solution:

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