Cho A, B là hai ma trận vuông cấp\[{\rm{n}} \ge 2\]Trường hợp nào sau đây luôn đúng?

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

A. m = -10

B. m = 25

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

D. \[{\rm{m}} \ne 10\]

A. m > 1

B. \(m < \frac{1}{{\sqrt 3 }}\)

C. \[{\rm{m}} \ne 0\]

D. \[\frac{{ - 4}}{5} < {\rm{m}} < 0\]

A. m = 11

B. m = 15

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

D. m = -21