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#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 4 Algebra of Matrices Exercise Multiple Choice Questions Question 14 Maths Textbook Solution.

Answer: The correct option is (C).

Given:$A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { and } \mathrm{A}^{2}=\mathrm{I}$

Solution: \begin{aligned} &A^{2}=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \times\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \\ &A^{2}=\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right] \\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}

$\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$

So,\begin{aligned} &\alpha^{2}+\beta \gamma=1 \\ &\alpha^{2}+\beta \gamma-1=0 \\ &1-\alpha^{2}-\beta \gamma=0 \end{aligned}

Hence, the correct option is (C).