Bt - Hang Cua He Vecto - Kgvt Con

  • November 2019
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BAØI TAÄP HAÏNG CUÛA HEÄ VEÙC TÔ – KGVT CON -----*----BAØI 1 : Tìm caùc heä veùc tô ÑLTT cöïc ñaïi cuûa heä veùc tô sau : a / u1   1, 3, 1

, u2   2,1, 7  ,

u3   1,9, 5 

b / u1   1, 4,1

, u2   0, 1,3 ,

u3   2,5, 4 

c / u1   1, 0, 1

, u2   0,1, 1 ,

u3   1, 1, 0 

d / u1   2,1, 0 

, u2   0, 2,1 ,

u3   2, 1, 2 

e / u1   1, 1, 0  , u2   2, 1, 1 , u3   0,1, 1 BAØI 2 : Tìm haïng cuûa caùc veùc tô sau :

, u4   2, 0, 2 

a / u1   1, 2, 1, 0 

, u2   0, 1,3,1

,

u3   3, 0, 2, 1

b / u1   2,3, 1

, u2   3, 1,5 

,

u3   1, 4,3

c / u1   1, 0, 2,3,1

, u2   2,1,3, 2, 0  ,

u3   3, 2,5,1, 4 

d / u1   1, 2, 0,1

, u2   1, 2,3, 1

,

u3   0, 4,3, 0 

e / u1   1, 4,8,12 

, u2   2,1,3,1

,

u3   2,8,16, 24 

, u4   0,1, 4, 6, 5 

,

u4   1,1, 2, 3

BAØI 3 : Xeùt xem taäp V coù phaûi laø khoâng gian veùc tô con cuûa ¡ 3 hay khoâng ? a / V    x1 , x2 , x3   ¡ 3 / x2  7 x1

b / V    x1 , x2 , x3   ¡ 3 / x1  x2  3 x3  2 c / V    x1 , x2 , x3   ¡ 3 / x2  0

d / V    x1 , x2 , x3   ¡ 3 / 2 x1  x3  3x3 

e / V    x1 , x2 , x3   ¡ 3 / x1  x2  2 x3  0 f / V    x1 , x2 , x3   ¡ 3 / x12  x2 

g / V    x1 , x2 , x3   ¡ 3 / x1  x2  x3  3   x1 , x2 , x3   ¡ / x1  x2  x3  0  h/ V    x1  2 x2  3 x3  0  

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