write all the formulae of exercise 2.1&2.2
Answers
Answer:
Let the cost of 1 kg onion , 1 kg wheat and 1 kg of rice be Rsx,Rsy,Rsz respectively.
So, 4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70
These equations can be written as
AX=B
where A=
⎣
⎢
⎢
⎡
4
2
6
3
4
2
2
6
3
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
,B=
⎣
⎢
⎢
⎡
60
90
70
⎦
⎥
⎥
⎤
Here,
∣A∣=4(12−12)−3(6−36)+2(4−24)
⇒∣A∣=90−40=50
Since, ∣A∣
=0
Hence, the system of equations is consistent and has a unique solution given by X==A
−1
B
A
−1
=
∣A∣
adjA
and adjA=C
T
C
11
=(−1)
1+1
∣
∣
∣
∣
∣
∣
4
2
6
3
∣
∣
∣
∣
∣
∣
⇒C
11
=12−12=0
C
12
=(−1)
1+2
∣
∣
∣
∣
∣
∣
2
6
6
3
∣
∣
∣
∣
∣
∣
⇒C
12
=−(6−36)=30
C
13
=(−1)
1+3
∣
∣
∣
∣
∣
∣
2
6
4
2
∣
∣
∣
∣
∣
∣
⇒C
13
=4−24=−20
C
21
=(−1)
2+1
∣
∣
∣
∣
∣
∣
3
2
2
3
∣
∣
∣
∣
∣
∣
⇒C
21
=−(9−4)=−5
C
22
=(−1)
2+2
∣
∣
∣
∣
∣
∣
4
6
2
3
∣
∣
∣
∣
∣
∣
⇒C
22
=12−12=0
C
23
=(−1)
2+3
∣
∣
∣
∣
∣
∣
4
6
3
2
∣
∣
∣
∣
∣
∣
⇒C
23
=−(8−18)=10
C
31
=(−1)
3+1
∣
∣
∣
∣
∣
∣
3
4
2
6
∣
∣
∣
∣
∣
∣
⇒C
31
=18−8=10
C
32
=(−1)
3+2
∣
∣
∣
∣
∣
∣
4
2
2
6
∣
∣
∣
∣
∣
∣
⇒C
32
=−(24−4)=−20
C
33
=(−1)
3+3
∣
∣
∣
∣
∣
∣
4
2
3
4
∣
∣
∣
∣
∣
∣
⇒C
33
=16−6=10
Hence, the co-factor matrix is C=
⎣
⎢
⎢
⎡
0
−5
10
30
0
−20
−20
10
10
⎦
⎥
⎥
⎤
⇒adjA=C
T
=
⎣
⎢
⎢
⎡
0
30
−20
−5
0
10
10
−20
10
⎦
⎥
⎥
⎤
⇒A
−1
=
∣A∣
adjA
=
50
1
⎣
⎢
⎢
⎡
0
30
−20
−5
0
10
10
−20
10
⎦
⎥
⎥
⎤
Solution is given by
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
50
1
⎣
⎢
⎢
⎡
0
30
−20
−5
0
10
10
−20
10
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
60
90
70
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
50
1
⎣
⎢
⎢
⎡
0−450+700
1800+0−1400
−1200+900+700
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
50
1
⎣
⎢
⎢
⎡
250
400
400
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
5
8
8
⎦
⎥
⎥
⎤
Hence, x=5,y=8,z=8
So, the cost of 1 kg onion is Rs 5 , 1 kg wheat is Rs 8 , 1 kg rice is Rs 8.
Answer:
Solutions 16 Questions (6 Long Answer Questions, 10 Short Answer Questions)
Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions)
Exercise 2.4 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions)
Exercise 2.5 Solutions 10 Questions (1 Long Answer Questions, 9 Short Answer Questions)
Exercise 2.6 Solutions 7 Questions (1 Long Answer Questions, 6 Short Answer Questions)
Access Answers of Maths NCERT Class 8 Chapter 2- Linear Equations in One Variable Exercise 2.1 Page Number 23
Solve the following equations.
1. x – 2 = 7
Solution:
x – 2 = 7
x=7+2
x=9
2. y + 3 = 10
Solution:
y + 3 = 10
y = 10 –3
y = 7
3. 6 = z + 2
Solution:
6 = z + 2
z + 2 = 6
z = 6-2
z=4
4. 3/7 + x = 17/7
Solution:
3/7 + x = 17/7
x = 17/7 – 3/7
x = 14/7
x = 2
5. 6x = 12
Solution:
6x = 12
x = 12/6
x = 2
6. t/5 = 10
Solution:
t/5 = 10
t = 10 × 5
t = 50
7. 2x/3 = 18
Solution:
2x/3 = 18
2x = 18 × 3
2x = 54
x = 54/2
x = 27
8. 1.6 = y/15
Solution:
1.6 = y/1.5
y/1.5 = 1.6
y = 1.6 × 1.5
y = 2.4
9. 7x – 9 = 16
Solution:
7x – 9 = 16
7x = 16+9
7x = 25
x = 25/7
10. 14y – 8 = 13
Solution:
14y – 8 = 13
14y = 13+8
14y = 21
y = 21/14
y = 3/2
11. 17 + 6p = 9
Solution:
17 + 6p = 9
6p = 9 – 17
6p = -8
p = -8/6
p = -4/3
12. x/3 + 1 = 7/15
Solution:
x/3 + 1 = 7/15
x/3 = 7/15 – 1
x/3 = (7 -15)/15
x/3 = -8/15
x = -8/15 × 3
x = -8/5