Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

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 (integer) 100

Solution

<p>$\because$ $$X = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right]$$</p> <p>$\therefore$ $${X^2} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$</p> <p>$\therefore$ $$Y = \alpha I + \beta X + \gamma {X^2}\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]$$</p> <p>$\because$ $Y\,.\,{Y^{ - 1}} = I$</p> <p>$\therefore$ $$\left[ {\matrix{ \alpha & \beta & \gamma \cr 0 & \alpha & \beta \cr 0 & 0 & \alpha \cr } } \right]\left[ {\matrix{ {{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr 0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr 0 & 0 & {{1 \over 5}} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$</p> <p>$\therefore$ $$\left[ {\matrix{ {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} & {{{\alpha - 2\beta + \gamma } \over 5}} \cr 0 & {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} \cr 0 & 0 & {{\alpha \over 5}} \cr } } \right] = \left[ {\matrix{ 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$</p> <p>$\therefore$ $\alpha$ = 5, $\beta$ = 10, $\gamma$ =15</p> <p>$\therefore$ ($\alpha$ $-$ $\beta$ + $\gamma$)<sup>2</sup> = 100</p>

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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