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When any Symmetric Matrix\(A\)is diagonalized orthogonallyas \(PD{P^{ - 1}}\), we have:

\(\begin{array}{l}{{\rm{x}}^T}A{\rm{x}} = {{\rm{y}}^T}D{\rm{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as x}} = Py} \right\}\\\\\left\| {\rm{x}} \right\| = \left\| {P{\rm{y}}} \right\| = \left\| {\rm{y}} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall {\rm{y}} \in \mathbb{R}} \right\}\end{array}\)

As per the question, we have:

The matrix\(A\)has eigenvalue 3 with eigenvector\({\rm{x}}\)which is a unit vector, then:

\(\begin{array}{c}A{\rm{x}} = 3{\rm{x}}\\{{\rm{x}}^T}A{\rm{x}} = {{\rm{x}}^T}3{\rm{x}}\\ = 3\left( {{{\rm{x}}^T}{\rm{x}}} \right)\\ = 3{\left\| {\rm{x}} \right\|^2}\\ = 3\end{array}\)

Hence, the required value is \({{\rm{x}}^T}A{\rm{x}} = 3\).