Step by Step Solutions

[Solution] Suppose that

Question: Suppose that

\[Q_{B}(\mathrm{x})=\mathrm{x}^{t}\left(\begin{array}{cc} 2 & -2 \\ -2 & -1 \end{array}\right) \mathrm{x}, \quad \text { for } \mathrm{x} \in \mathbb{R}^{2}\]

Find a diagonal matrix \(D\) and orthogonal matrix \(P\) so that \(\left(\begin{array}{cc}2 & -2 \\ -2 & -1\end{array}\right)=P D P^{t}\). For \(\mathrm{x} \in \mathbb{R}^{2}\), let \(\mathbf{y}\) be given by \(\mathrm{x}=P \mathbf{y}\), so that \(Q_{B}(\mathrm{x})=Q_{D}(\mathbf{y})\). Use this calculation to describe the shape of the curve in the plane given by the equation

\[2 x_{1}^{2}-4 x_{1} x_{2}-x_{2}^{2}=-8\]

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