JEE Advance - Mathematics (2015 - Paper 1 Offline - No. 18)

Which of the following values of $$\alpha$$ satisfy the equation

$$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {{{(2 + \alpha )}^2}} & {{{(2 + 2\alpha )}^2}} & {{{(2 + 3\alpha )}^2}} \cr {{{(3 + \alpha )}^2}} & {{{(3 + 2\alpha )}^2}} & {{{(3 + 3\alpha )}^2}} \cr } } \right| = - 648\alpha $$ ?

$$-$$4
9
$$-$$9
4

Explanation

$$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {{{(2 + \alpha )}^2}} & {{{(2 + 2\alpha )}^2}} & {{{(2 + 3\alpha )}^2}} \cr {{{(3 + \alpha )}^2}} & {{{(3 + 2\alpha )}^2}} & {{{(3 + 3\alpha )}^2}} \cr } } \right| = - 648\alpha $$

Applying $${R_3} \to {R_3} - {R_2},{R_2} \to {R_2},{R_1}$$, we get

$$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr {5 + 2\alpha } & {5 + 4\alpha } & {5 + 6\alpha } \cr } } \right| = - 648\alpha $$

Applying $${R_3} \to {R_3} - {R_2}$$, we get

$$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr 2 & 2 & 2 \cr } } \right| = - 648\alpha $$

Applying $${C_3} \to {C_3} - {C_2},{C_2} \to {C_2} - {C_1}$$, we get

$$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {2\alpha } & {2\alpha } \cr 2 & 0 & 0 \cr } } \right| = - 648\alpha $$

Expanding, $$2{\alpha ^2}(3\alpha + 2) - 2{\alpha ^2}(5\alpha + 2) = - 324\alpha $$

$$ \Rightarrow - 4{\alpha ^3} = - 324\alpha \Rightarrow \alpha ({\alpha ^2} - 81) = 0$$

$$\therefore$$ $$\alpha = 0, - 9,9$$

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