Question

If $\left| \begin{matrix} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{matrix} \right| = 0$and $\alpha \neq \frac{1}{2},$ then.

• A. $a,b,c$ are in A. P.
• B. $a,b,c$are in G. P.
• C. $a,b,c$are in H. P.
• D. None of these

Answer 1

Answer: (B)

$a,b,c$are in G. P.

Answer 2

$\left| \begin{matrix} \mathbf{a} & \mathbf{b} & \mathbf{a\alpha}\mathbf{-}\mathbf{b} \\ \mathbf{b} & \mathbf{c} & \mathbf{b\alpha}\mathbf{-}\mathbf{c} \\ \mathbf{2} & \mathbf{1} & \mathbf{0} \end{matrix} \right|\mathbf{= 0}$

⇒ $a\lbrack - (b\alpha - c)\rbrack - b\lbrack - 2(b\alpha - c)\rbrack + \lbrack a\alpha - b)(b - 2c)\rbrack = 0$

⇒$- ab\alpha + ac + 2b^{2}\alpha - 2bc + ab\alpha - 2ac\alpha - b^{2} + 2bc = 0$

⇒ $ac + 2b^{2}\alpha - 2ac\alpha - b^{2} = 0$

⇒ $(ac - b^{2}) - 2\alpha(ac - b^{2}) = 0$

⇒ $ac - b^{2} = 0$or $1 - 2\alpha = 0$ $\Rightarrow$ $b^{2} = ac$or $\alpha = \frac{1}{2}$

$\because\alpha \neq \frac{1}{2}$ (As given in question)

So, $b^{2} = ac$ i.e, $a,b,c$are in G.P.