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
-9

$\left|\begin{array}{ccc}(1+\alpha {)}^2& (1+2\alpha {)}^2& (1+3\alpha {)}^2\\ (2+\alpha {)}^2& (2+2\alpha {)}^2& (2+3\alpha {)}^2\\ (3+\alpha {)}^2& (3+2\alpha {)}^2& (3+3\alpha {)}^2\end{array}\right|$
$C_3\to C_3-C_2$ and $C_2\to C_2-C_1$, above determinant reduces to:
$\Rightarrow \alpha \left|\begin{array}{ccc}(1+\alpha {)}^2& 2+3\alpha & 2+5\alpha \\ (2+\alpha {)}^2& 4+3\alpha & 4+5\alpha \\ (3+\alpha {)}^2& 6+3\alpha & 6+5\alpha \end{array}\right|$
$C_3\to C_3-C_2$,
$\Rightarrow \alpha \left|\begin{array}{ccc}(1+\alpha {)}^2& 2+3\alpha & 2\alpha \\ (2+\alpha {)}^2& 4+3\alpha & 2\alpha \\ (3+\alpha {)}^2& 6+3\alpha & 2\alpha \end{array}\right|\Rightarrow 2{\alpha }^2\left|\begin{array}{ccc}(1+\alpha {)}^2& 2+3\alpha & 1\\ (2+\alpha {)}^2& 4+3\alpha & 1\\ (3+\alpha {)}^2& 6+3\alpha & 1\end{array}\right|$
$R_1\to R_1-R_2$ and $R_2\to R_2-R_3$,
$\Rightarrow 2{\alpha }^2\left|\begin{array}{ccc}(1+\alpha {)}^2& 2+3\alpha & 0\\ (2+\alpha {)}^2& 4+3\alpha & 0\\ (3+\alpha {)}^2& 6+3\alpha & 1\end{array}\right|=-8{\alpha }^3$
$\Rightarrow -8{\alpha }^3=-648\alpha$, which gives $\alpha =\pm 9$
Hence, option B,C.