Question

Which of the following values of $\alpha$ satisfy the equation
$\left|\begin{array}{ccc}(1+\alpha {)}^2& (1+2\alpha {)}^2& (1+3\alpha {)}^2\\ (2+\alpha {)}^2& (2+2\alpha {)}^2& (2+3\alpha {)}^2\\ (3+\alpha {)}^2& (3+2\alpha {)}^2& (3+3\alpha {)}^2\end{array}\right|=-648\alpha$?

• A. -4
• B. 9
• C. -9
• D. 4

Answer 1

Answer: (B), (C)

-9

Given determinant could be expressed as product of two determinants.
i.e. $\left|\begin{array}{ccc}(1+\alpha {)}^2& (1+2\alpha {)}^2& (1+3\alpha {)}^2\\ (2+\alpha {)}^2& (2+2\alpha {)}^2& (2+3\alpha {)}^2\\ (3+\alpha {)}^2& (3+2\alpha {)}^2& (3+3\alpha {)}^2\end{array}\right|=-648\alpha$
$\Rightarrow \left|\begin{array}{ccc}1+2\alpha +{\alpha }^2& 1+4\alpha +4{\alpha }^2& 1+6\alpha +9{\alpha }^2\\ 4+4\alpha +{\alpha }^2& 4+8\alpha +4{\alpha }^2& 4+12\alpha +9{\alpha }^2\\ 9+6\alpha +{\alpha }^2& 9+12\alpha +4{\alpha }^2& 9+18\alpha +9{\alpha }^2\end{array}\right|=-648\alpha$
$\Rightarrow \left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ 4& 2\alpha & {\alpha }^2\\ 9& 3\alpha & {\alpha }^2\end{array}\right|.\left|\begin{array}{ccc}1& 1& 1\\ 2& 4& 6\\ 1& 4& 9\end{array}\right|=-648\alpha \Rightarrow {\alpha }^3\left|\begin{array}{ccc}1& 1& 1\\ 4& 2& 1\\ 9& 3& 1\end{array}\right|.\left|\begin{array}{ccc}1& 1& 1\\ 2& 4& 6\\ 1& 4& 9\end{array}\right|=-648\alpha$
$\Rightarrow -8{\alpha }^3=-648\alpha \Rightarrow {\alpha }^3-81\alpha =0\Rightarrow \alpha ({\alpha }^2-81)=0\therefore \alpha =0,\pm 9$