Let \(\mathrm{A}=\left[\begin{array}{cc}x & 2 y \\ -1 & y\end{array}\right]\); \(x, y \in \mathbf{R}\).
If \(\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{ll}1 & 0 \\ 0 & \alpha\end{array}\right]\) \((\alpha \in \mathbf{R})\), then \(\alpha+y^{2}\) is equal to :
If \(\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{ll}1 & 0 \\ 0 & \alpha\end{array}\right]\) \((\alpha \in \mathbf{R})\), then \(\alpha+y^{2}\) is equal to :
Show Hint
- \(\sqrt{2}-1\)
- 2
- \(\sqrt{2}\)
- 1
The Correct Option is C
Solution and Explanation
Equating with \(\begin{bmatrix}1&0\\0&\alpha\end{bmatrix}\): \(x=2y^2\), \(x^2+4y^2=1\), \(\alpha=1+y^2\).
Substituting: \(4y^4+4y^2-1=0\), so \(y^2=\frac{-4+\sqrt{16+16}}{8}=\frac{\sqrt2-1}{2}\) (taking the positive root).
Then \(\alpha+y^2=1+2y^2=1+(\sqrt2-1)=\sqrt2\).
Answer: (3) \(\sqrt{2}\)
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