1. CH働NG I. MA TR畉N 畛NH TH畛C H畛 PH働NG
TRNH TUY畉N TNH
則1. KHI NI畛M V畛 MA TR畉N
Bi 1: Th畛c hi畛n c叩c ph辿p t鱈nh sau
1 3 1 4
1 1 0 2
6 5 2 11 5 1 2 3 2 1
1. 4. 0 1 1 0
0 0 7 3 2 3 0 4 1 0 2 1 3 2
4
2 3 3
4 cos sin
n
2 1 1 5. (n , 0 < 2 )
2. 2 (1 2) sin cos
1 2 1
0
1 an
0 1 , a R v n
3.
2 1
Bi 2: Cho A = v f ( x ) = 3x + 2 x 4 . T鱈nh f ( A) .
2
0 3
Bi 3:
x y x 6 4 x + y
1. T狸m c叩c s畛 th畛c x, y , z , w sao cho 3 = + .
z w 1 2 w z + w 3
2 1
2. T狸m t畉t c畉 c叩c ma tr畉n c畉p 2 giao ho叩n v畛i ma tr畉n A = .
0 1
1 1 3 2 2
2 1 2
Bi 4: Cho c叩c ma tr畉n A = 1 2 2 , B = 1 2 , C = . T鱈nh C B A .
t t t
2 2 5 3 2 2 3 1
Bi 5: T狸m ma tr畉n X trong c叩c tr動畛ng h畛p sau
2 1 3 2 1 2
1. . X. =
4 5 5 3 3 4
2 1 1 1 1 1
2. . X X . =
1 2 1 1 1 1
1 2 2 3 5 1 5
3. 2 5 4 X - 7 6 = 3 2 2
2 4 5 2 1 2 1
則2. 畛NH TH畛C
2. Bi 6: T鱈nh c叩c 畛nh th畛c sau 但y
7 6 5 2 3 4 1 2 3 4
1. 1 2 1 2. 5 6 7 2 3 4 1
3.
3 2 2 8 9 1 3 4 1 4
4 1 2 3
1 2 3 4 a+x x x x2 + 1 xy xz
2 3 4 1 5. x b+x x 6. xy y +1
2
yz
4.
3 4 1 4 x x c+x xz yz z +1
2
4 1 2 3
Bi 7: T鱈nh c叩c 畛nh th畛c c畉p n sau 但y
1 2 3 ....... n 1 n 1 2 3 ....... n 1 n
1 0 3 ....... n 1 n 2 2 3 ....... n 1 n
1 2 0 ........ n 1 n 3 3 3 ....... n 1 n
1. 3.
...................................... .... ..... ..... ....... ....... ...
1 2 3 ............ 0 n n 1 n 1 n 1 ....... n 1 n
1 2 3 ........ n 1 0 n n n ....... n n
1 1 1 ....... 1 1
x a a a
1 2 2 ....... 2 2
a x a a
1 2 3 ....... 3 3
3. 4. a a x a
... ... ... ....... .... ....
1 2 3 ....... n 1 n 1
a a a x
1 2 3 ....... n 1 n
Bi 8:Gi畉i c叩c ph動董ng tr狸nh sau 但y
1 x x 2 x3
x x +1 x + 2
1 2 4 8
1. =0 2. x + 3 x + 4 x + 5 = 0
1 3 9 27
x+6 x+7 x+8
1 4 16 64
則3. H畉NG C畛A MA TR畉N
Bi 9: T狸m h畉ng c畛a c叩c ma tr畉n sau
3. 1 3 5 1
2 1 3 2 4
1. 2 1 3 4
2. 4 2 5 1 7
5 1 1 7 2 1
1 8 2
7 7 9 1
0 2 4
2 4 3 1 0
1 4 5
1 2 1 4
2
3. 3 1 7 3.
0 1 1 3 1
0 5 10
2 3 1 7 4 4 5
0
Bi 10: T湛y theo tham s畛 m , h達y t狸m h畉ng c畛a c叩c ma tr畉n sau
1 2 3 3 1 1 4
1. 4 5 6 2.
2 2 4 3
7 8 9 m 4 10 1
10 m 12 1 7 17 3
m 1 1 1
Bi 11: Cho ma tr畉n A = 1 m 1 m . T狸m m 畛 r ( A) < 3 .
1 1 1 m2
則4. MA TR畉N NGH畛CH 畉O
Bi 12: T狸m ma tr畉n ngh畛ch 畉o c畛a c叩c ma tr畉n sau 但y b畉ng ph動董ng ph叩p bi畉n 畛i s董
c畉p
1 0 3 1 3 2 1 3 5
1. 2 1 1 2. 2 1 3 3. 5 0 1
3 2 2 3 2 1 3 1 0
1 2 0 1 2 1 0 2 1 1 0 0
4. 1 1 2 0 5. 2 2 1 0 6. 0 1 1 0
0 1 1 2 0 2 2 1 0 0 1 1
2 0 1 1 1 0 2 2 1 0 0 1
則5. H畛 PH働NG TRNH TUY畉N TNH
Bi 13: Gi畉i c叩c h畛 ph動董ng tr狸nh tuy畉n t鱈nh sau
ｧ x1 2x 2 + x 3 + 2x 4 = 1 ｧ x1 + 3x 2 + 5 x3 + x 4 = 0
ｪ ｪ
1. ｨ x 1 + x 2 x 3 + x 4 = 2 2. ｨ 4 x1 7 x 2 3x3 x 4 = 0
ｪx + 7x 5x x = 0 ｪ3x + 2 x + 7 x + 8 x = 0
ｩ 1 2 3 4 ｩ 1 2 3 4
Bi 14: Gi畉i v bi畛n lu畉n c叩c h畛 ph動董ng tr狸nh sau
4. ｧmx + y +z = 1 ｧ x + y + (1 m )z = m + 2
ｪ ｪ
1. ｨ x + my + z = m 2. ｨ(1 + m )x y +2 z = 0
ｪ x + y + mz = m 2 ｪ 2x my +3z = m+2
ｩ ｩ
ｧ x1 +3x 2 +2 x 3 +4 x 4 = 1
ｧx 1 2 x 2 +x 3 +2 x 4 = 1 ｪ2 x
ｪ ｪ 1 +5x 2 +2 x 3 +9 x 4 = 1
3. ｨx 1 +x 2 x 3 +x 4 = m 4. ｨ
ｪx +7x 2 5x 3 x 4 = 4m ｪ x1 +5x 2 +6 x 3 + mx 4 = 3
ｩ 1 ｪ x1 +3x 2 +4 x 3 +3x 4 = 2
ｩ
ｧ x1 +2 x 2 x3 = 2
ｧ 2x1 x 2 + x 3 + x 4 = 1 ｪ2 x
ｪ ｪ 1 3x 2 +7x 3 = 1
5. ｨ x 1 + 2 x 2 x 3 + 4 x 4 = 2 6. ｨ
ｪx + 7x 4 x + 11x = m ｪ x 1 +x 2 +3x 3 = 6
ｩ 1 2 3 4
ｪ 5x 1 +x 2 +2 x 3 = m
ｩ
ｧ x1 +2 x 2 +3x 3 + mx 4 = m+2
ｪx +x 2 +x 3 + mx 4 = m +1
ｪ 1
ｪ
7. ｨ2 x 1 +3x 2 +4 x 3 +2 mx 4 = 2m + 3
ｪ3 x +4 x 2 +2 x 3 +3mx 4 = 3m + 1
ｪ 1
ｪ x1
ｩ +x 2 +2 x 3 +2 mx 4 = m2 + m + 2
ｧmx +y +z = m
ｪ
Bi 15: Cho h畛 ph動董ng tr狸nh ｨ 2 x + (1 + m )y + (1 + m )z = m 1 . T狸m tham s畛 m
ｪ x +y + mz = 1
ｩ
畛 h畛 ph動董ng tr狸nh tr棚n c坦 nghi畛m.
ｧax 3y + z = 2
ｪ
Bi 16: Cho h畛 ph動董ng tr狸nh ｨax + y +2 z = 3 (I), trong 坦 a, b l tham s畛.
ｪ3x +2 y + z = b
ｩ
1. X叩c 畛nh a, b 畛 h畛 (I) l h畛 Cramer. Khi 坦 h達y t狸m nghi畛m c畛a h畛 theo a, b .
2. T狸m a, b 畛 h畛 (I) v担 nghi畛m.
3. T狸m a, b 畛 h畛 (I) c坦 v担 s畛 nghi畛m v t狸m nghi畛m t畛ng qu叩t c畛a h畛.
ｧ x 1 3x 2 + 4 x 3 x 4 = 0
ｪ
ｪ2 x1 + x 2 2 x 3 + 2 x 4 = 0
Bi 17: T狸m h畛 nghi畛m c董 b畉n c畛a h畛 ph動董ng tr狸nh ｨ .
ｪ3x1 2 x 2 + 2 x 3 + x 4 = 0
ｪ x 1 + 4 x 2 6 x 3 + 3x 4 = 0
ｩ
