The cost of 12kg of oranges and 24kg of onions is 21600=. The cost of 24kg
of oranges and 12kg of onions is 18000=.
a. Find the cost of one kg of onions
b. Find the cost one kg of oranges
c. What is the cost of 5kg of oranges and 10kg o
Answers
Answer:
Step-by-step explanation:
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.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.