MCQ
Given that $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ and $A^2=31$, then
- A$1+\alpha^2+\beta \gamma=0$
- B$1-\alpha^2-\beta \gamma=0$
- C$3-\alpha^2-\beta \gamma=0$
- D$3+\alpha^2+\beta \gamma=0$
✓
Answer
We have,
\[\begin{array}{l}
\text { 8. (c) : We have, } A=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
\Rightarrow A^2=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2
\end{array}\right]
\end{array}\]
But $A^2=31$
\[\begin{array}{l}
\Rightarrow\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \alpha^2+\beta \gamma
\end{array}\right]=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
\Rightarrow \alpha^2+\beta \gamma=3 \\
\Rightarrow 3-\alpha^2-\beta \gamma=0
\end{array}\]
\[\begin{array}{l}
\text { 8. (c) : We have, } A=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
\Rightarrow A^2=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2
\end{array}\right]
\end{array}\]
But $A^2=31$
\[\begin{array}{l}
\Rightarrow\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \alpha^2+\beta \gamma
\end{array}\right]=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
\Rightarrow \alpha^2+\beta \gamma=3 \\
\Rightarrow 3-\alpha^2-\beta \gamma=0
\end{array}\]
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