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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices1 Mark
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
    $1-\alpha^2-\beta \gamma=0$
  • ✓
    $3-\alpha^2-\beta \gamma=0$
  • D
    $3+\alpha^2+\beta \gamma=0$
✓

Answer

Correct option: C.
$3-\alpha^2-\beta \gamma=0$
We have,
We have, $ A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] $
$\Rightarrow A^2=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right]=\left[\begin{array}{cc} \alpha^2+\beta \gamma & 0 \\
0 & \gamma \beta+\alpha^2 \end{array}\right]$
But $A^2=31$
$\Rightarrow\left[\begin{array}{cc} \alpha^2+\beta \gamma & 0 \\ 0 & \alpha^2+\beta \gamma \end{array}\right]=\left[\begin{array}{ll} 3 & 0 \\  0 & 3\end{array}\right] \\
\Rightarrow \alpha^2+\beta \gamma=3$
$\Rightarrow 3-\alpha^2-\beta \gamma=0$

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