91Ó°ÊÓ

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

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

As per the question, we have:

\(3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{} = 7y_1^2 + 4y_2^2\)

The matrix of thequadratic formwill be:

\(A = \left[ {\begin{array}{*{20}{c}}3&3&1\\3&3&1\\1&1&5\end{array}} \right]\)

And for each eigenvalue of\(A\)defined in the right-hand side of the equation will be expressed as:

\(\begin{array}{l}{\lambda _1} = 7 \Rightarrow {P_1} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\end{array}} \right]\\{\lambda _2} = 4 \Rightarrow {P_2} = \left[ {\begin{array}{*{20}{c}}{ - \frac{1}{{\sqrt 6 }}}\\{ - \frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\end{array}} \right]\\{\lambda _3} = 0 \Rightarrow {P_3} = \left[ {\begin{array}{*{20}{c}}{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\0\end{array}} \right]\end{array}\)

Hence,therequired change of variable is:

\(P = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 6 }}}&{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{2}{{\sqrt 6 }}}&0\end{array}} \right]\).