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JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 18)

Let $$X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right],\,Y = \alpha I + \beta X + \gamma {X^2}$$ and $$Z = {\alpha ^2}I - \alpha \beta X + ({\beta ^2} - \alpha \gamma ){X^2}$$, $$\alpha$$, $$\beta$$, $$\gamma$$ $$\in$$ R. If $${Y^{ - 1}} = \left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right]$$, then ($$\alpha$$ $$-$$ $$\beta$$ + $$\gamma$$)2 is equal to ____________.
Answer
100

Explanation

$$\because$$ $$X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right]$$

$$\therefore$$ $${X^2} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$

$$\therefore$$ $$Y = \alpha I + \beta X + \gamma {X^2}\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]$$

$$\because$$ $$Y\,.\,{Y^{ - 1}} = I$$

$$\therefore$$ $$\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]\left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$\therefore$$ $$\left[ {\matrix{ {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} & {{{\alpha - 2\beta + \gamma } \over 5}} \cr 0 & {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} \cr 0 & 0 & {{\alpha \over 5}} \cr } } \right] = \left[ {\matrix{ 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$\therefore$$ $$\alpha$$ = 5, $$\beta$$ = 10, $$\gamma$$ =15

$$\therefore$$ ($$\alpha$$ $$-$$ $$\beta$$ + $$\gamma$$)2 = 100

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