Question
If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then:
  1. a, b, c are in H.P.
  2. $\alpha$ is a root of 4ax2 + 12bx + 9c = 0 or a, b, c are in G.P.
  3. a, b, c are in G.P. only.
  4. a, b, c are in A.P.

Answer

  1. $\alpha$ is a root of 4ax2 + 12bx + 9c = 0 or a, b, c are in G.P.

Solution:

Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$ [Applying C1 → C1 - C2]

$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$

$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c}&\text{c}\end{vmatrix}$ [Expanding along R3]

$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$

But $\triangle=0$ [Given]

$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$

$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$

Or $(\text{ac}-\text{b}^2)=0$

$\Rightarrow\alpha$ is a root of 4ax2 + 12bx + 9c = 0

Or ac = b2, i.e. a, b, c are in G.P.

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