Giải hệ phương trình tuyến tính \[\left\{ \begin{array}{l}4{x_1} + 3{x_2} + {x_3} + 5{x_4} = 7\\{x_1} - 2{x_2} - 2{x_3} - 3{x_4} = 3\\3{x_1} - {x_2} + 2{x_3} = - 1\\2{x_1} + 3{x_2} + 2{x_3} - 8{x_4} = - 7\end{array} \right.\]

A. u = (1,−2,1),v = (2,1,−1)

B. u = (2,−3,13),v = (2,0,8),w = (8,−1,8),x = (3,−9,7)

C. u = (1,1,1),v = (1,2,3),w = (2,−1,1)

D. u = (1,1,2),v = (1,2,5),w = (5,3,4)

A. λ=10

B. λ=−11

C. λ=12

D. λ=11

A. u = (1,−2,1),v = (2,1,−1),w = (7,−1,4)

B. u = (1,−3,7),v = (2,0,8),w = (8,−1,8),x(3,−9,7)

C. u = (1,2,−3),v = (1,−3,2),w = (2,−1,5)

D. u = (2,−3,13),v = (0,0,0),w = (1,−10,11)

A. u = (2,1,−3),v = (3,2,−5),w(1,−1,1)

B. u = (2,−1,3),v = (4,1,2),w = (8,−1,8)

C. u = (3,1,4),v = (2,−3,5),w = (5,−2,9),s=(1,4,−1)

D. u = (3,1,13),v = (2,7,4),w = (1,−10,11)

A. Các véc tơ có dạng (x,0,z)

B. Các véc tơ có dạng (x, y,1)

C. Các véc tơ có dạng (x,y,z) thoả mãn x + y + z = 0

D. Các véc tơ có dạng (x,y,z), 2x-y+z=0, x+y-4z=0

A. r = 4

B. r = 2

C. r = 3

D. r = 1

A. \[\left\{ \begin{array}{l}\left( {x,y,z} \right) + \left( {x\prime ,y\prime ,x\prime } \right) = (x + x\prime ,y + y\prime ,z + z\prime \\\alpha \left( {x,y,z} \right) = \left( {\alpha x,y,z} \right);\alpha \in \mathbb{R}\end{array} \right.\]

B. \[\left\{ \begin{array}{l}\left( {x,y,z} \right) + \left( {x\prime ,y\prime ,x\prime } \right) = (x + x\prime ,y + y\prime ,z + z\prime )\\\alpha \left( {x,y,z} \right) = \left( {2\alpha x,2\alpha y,2\alpha z} \right);\alpha \in \mathbb{R}\end{array} \right.\]

C. \[\left\{ \begin{array}{l}\left( {x,y,z} \right) + \left( {x\prime ,y\prime ,x\prime } \right) = (x + x\prime + 1,y + y\prime + 1,z + z\prime + 1)\\\alpha \left( {x,y,z} \right) = \left( {0,0,0} \right);\alpha \in \mathbb{R}\end{array} \right.\]

D. \[\left\{ \begin{array}{l}\left( {x,y,z} \right) + \left( {x\prime ,y\prime ,x\prime } \right) = (x + x\prime ,y + y\prime ,z + z\prime )\\\alpha \left( {x,y,z} \right) = \left( {\alpha x,\alpha y,\alpha z} \right);\alpha \in \mathbb{R}\end{array} \right.\]