Question

If the system of linear equations

$x + 2ay + az = 0,x + 3\text{by} + bz = 0,x + 4cy + cz = 0$ has a

non-zero solution, then a, b, c

• A. Are in A.P.
• B. Are in G.P.
• C. Are in H.P.
• D. Satisfy $a + 2b + 3c = 0$

Answer 1

Answer: (C)

Are in H.P.

Answer 2

System of linear equations has a non-zero solution, then $\left| \begin{matrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{matrix} \right| = 0$

Applying $C_{2} \rightarrow C_{2} - 2C_{3}$; $\left| \begin{matrix} 1 & 0 & a \\ 1 & b & b \\ 1 & 2c & c \end{matrix} \right| = 0$

Applying $R_{3} \rightarrow R_{3} - R_{2}\text{and}R_{2} \rightarrow R_{2} - R_{1}$

$\left| \begin{matrix} 1 & 0 & a \\ 0 & b & b - a \\ 0 & 2c - b & c - b \end{matrix} \right| = 0$ $\because x 51 = 100 x + 50 + 1$ $b(c - b) - (b - a)(2c - b) = 0$.

On simplification $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$; $\therefore$ a, b, c are in H.P.