For \(\alpha, \beta \in \mathbb{R}\), if the matrices \(\mathrm{A} = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}\), \(\mathrm{B} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}\) and \(\mathrm{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) satisfy the equation \((\mathrm{A} * \mathrm{B}) * 2\mathrm{I} = 20\mathrm{I}\), where \(*\) is defined as \(\mathrm{A} * \mathrm{B} = \mathrm{A}^2 + \mathrm{B}^2\), then \(|\alpha\beta|\) is equal to :
Show Hint
- \(2\sqrt{3}\)
- \(2\sqrt{2}\)
- 4
- 2
The Correct Option is D
Solution and Explanation
\(\mathrm{A}*\mathrm{B}=\mathrm{diag}(2\alpha^2,\ \alpha^2+\beta^2)\).
\((\mathrm{A}*\mathrm{B})*2\mathrm{I}=(\mathrm{A}*\mathrm{B})^2+(2\mathrm{I})^2=\mathrm{diag}(4\alpha^4+4,\ (\alpha^2+\beta^2)^2+4)=20\mathrm{I}\).
First entry: \(4\alpha^4=16\Rightarrow\alpha^2=2\).
Second entry: \((\alpha^2+\beta^2)^2=16\Rightarrow\alpha^2+\beta^2=4\Rightarrow\beta^2=2\).
\(|\alpha\beta|=\sqrt{\alpha^2\beta^2}=\sqrt{4}=2\).
Answer: (4) 2
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