Cho dạng toàn phương Q: R³ R có ma trận trong cơ sở chính tắc\[{\rm{A}} = \left( {\begin{array}{*{20}{c}}1&{\rm{m}}&{ - 1}\\{\rm{m}}&1&2\\{ - 1}&2&5\end{array}} \right)\].Với giá trị nào của tham số m thì dạng toàn phương Q, xác định dương:

A. \[\left\{ {{{\rm{v}}_1} = (2, - 1,0,0,0),{{\rm{v}}_2} = ( - 17,0,5,0,1),{{\rm{v}}_3} = (13,0, - 4,1,0)} \right\}\]

B. \[\left\{ {{{\rm{v}}_1} = (2, - 1,0,0,0),{{\rm{v}}_2} = ( - 17,0,5,0,1)} \right\}\]

C. \[\left\{ {{{\rm{v}}_1} = (2, - 1,0,0,0),{v_2} = (7,0,5,0,1),{{\rm{v}}_2} = (13,0, - 4,1,0)} \right\}\]

D. \[\left\{ {{{\rm{v}}_1} = (2, - 1,0,0,0),{v_2} = ( - 17,0,5,0,1),{{\rm{v}}_2} = (15,1, - 5,0, - 1)} \right\}\]

A. \[{{\rm{v}}_1} = (3,1,0, - 4),{{\rm{v}}_2} = (1, - 3,5,4)\]

B. \[{{\rm{v}}_1} = (4,1,0,6),{{\rm{v}}_2} = (2, - 1,3,0),{{\rm{v}}_3} = (1, - 1,3,2)\]

C. \[{{\rm{v}}_1} = (1,2,0),{{\rm{v}}_2} = (4,4,0,1)\]

D. \[{{\rm{v}}_1} = (2,4,2,0),{{\rm{v}}_2} = (5,6,1,2)\]

A. \[{{\rm{(}}{{\rm{W}}_{\rm{1}}}{\rm{ + }}{{\rm{W}}_{\rm{2}}}{\rm{)}}^ \bot }{\rm{ = }}{{\rm{W}}_{\rm{1}}}^ \bot \cap {{\rm{W}}_{\rm{2}}}^ \bot \]

B. \[{{\rm{(}}{{\rm{W}}_{\rm{1}}} \cap {\rm{ }}{{\rm{W}}_{\rm{2}}}{\rm{)}}^ \bot }{\rm{ = }}{{\rm{W}}_{\rm{1}}}^ \bot + {{\rm{W}}_{\rm{2}}}^ \bot \]

C. \[{{\rm{(}}{{\rm{W}}_{\rm{1}}} \cap {\rm{ }}{{\rm{W}}_{\rm{2}}}{\rm{)}}^ \bot }{\rm{ = }}{{\rm{W}}_{\rm{1}}}^ \bot \cup {{\rm{W}}_{\rm{2}}}^ \bot \]

D. \[{{\rm{(}}{{\rm{W}}_{\rm{1}}} \subset {\rm{ }}{{\rm{W}}_{\rm{2}}}{\rm{)}}^ \bot }{\rm{ = }}{{\rm{W}}_{\rm{1}}}^ \bot \supset {{\rm{W}}_{\rm{2}}}^ \bot \]

A. \[{{\rm{v}}_{\rm{1}}}{\rm{ = }}{{\rm{e}}_{\rm{1}}}{\rm{ = (1, 0),}}{{\rm{v}}_{\rm{2}}}{\rm{ = }}{{\rm{e}}_{\rm{2}}}{\rm{ = (0, 1)}}\]

B. \[{{\rm{v}}_{\rm{1}}}{\rm{ = }}{{\rm{e}}_{\rm{1}}}{\rm{ = (1, 0),}}{{\rm{v}}_{\rm{2}}}{\rm{ = (2, 1)}}\]

C. \[{{\rm{v}}_{\rm{1}}}{\rm{ = (1, 2),}}{{\rm{v}}_{\rm{2}}}{\rm{ = }}{{\rm{e}}_{\rm{2}}}{\rm{ = (2, 1)}}\]

A. m = 11

B. m = 15

C. \[{\rm{m}} \ne - 11\]

D. m = -21

A. Hệ vô nghiệm

B. \(\left\{ {\begin{array}{*{20}{c}}{m \ne 0,}\\{m = 0}\end{array}} \right. \Rightarrow {x_1} = \frac{{ - 5{x_3} - 13{x_4} - 3}}{2};{x_1} = \frac{{ - 7{x_3} - 19{x_4} - 7}}{2}\)

C. \(\left\{ {\begin{array}{*{20}{c}}{m = 9,}\\{m \ne 9}\end{array}} \right. \Rightarrow {x_1} = \frac{{2{x_1} + 11{x_2} - 3}}{2};{x_1} = \frac{{ - 5{x_1} + 21{x_2} - 7}}{2}\)