Tìm một cơ sở { v 1 , v 2 , v 3 } của R3 sao cho biểu thức toạ độ của Q trong cơ sở này có dạng chính tắc ( x , y , z ) = X v 1 + Y v 2 + Z v 3 ; Q ( x , y , z ) = α x 2 + β y 2 + γ z 2

\[{{\rm{v}}_1} = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right),{{\rm{v}}_2} = \left( {0,\frac{1}{{\sqrt 2 }},\frac{{ - 1}}{{\sqrt 2 }}} \right),{{\rm{v}}_3} = \left( {\frac{{ - 4}}{{\sqrt {18} }},\frac{1}{{\sqrt {18} }},\frac{1}{{\sqrt {18} }}} \right);{\rm{\alpha }} = 9,{\rm{\beta }} = 18,{\rm{\gamma }} = 18\]

\[{{\rm{v}}_1} = \left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right),{{\rm{v}}_2} = \left( {\frac{1}{{\sqrt 2 }},\frac{{ - 1}}{{\sqrt 2 }},0} \right),{{\rm{v}}_3} = \left( {\frac{1}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }},\frac{{ - 2}}{{\sqrt 6 }}} \right);{\rm{\alpha }} = 5,{\rm{\beta }} = 10,{\rm{\gamma }} = 10\]

\[\begin{array}{*{20}{l}}{{{\rm{v}}_1} = \left( {\frac{2}{3},\frac{2}{3},\frac{{ - 1}}{3}} \right),{{\rm{v}}_2} = \left( {\frac{1}{3},\frac{{ - 2}}{3},\frac{2}{3}} \right),{{\rm{v}}_3} = \left( {\frac{2}{3},\frac{1}{3},\frac{2}{3}} \right);{\rm{\alpha }} = 3,{\rm{\beta }} = 5,{\rm{\gamma }} = - 1}\\{{\rm{p}} = 1,{\rm{q}} = 2}\\{\left( {\begin{array}{*{20}{c}}2&{ - 6}\\0&1\end{array}} \right)}\end{array}\]

\[{{\rm{v}}_1} = \left( {\frac{2}{3},\frac{2}{3},\frac{{ - 1}}{3}} \right),{{\rm{v}}_2} = \left( {\frac{1}{3},\frac{{ - 2}}{3},\frac{2}{3}} \right),{{\rm{v}}_3} = \left( {\frac{2}{3},\frac{1}{3},\frac{2}{3}} \right);{\rm{\alpha }} = 1,{\rm{\beta }} = 1,{\rm{\gamma }} = 2\]