Math, asked by kushisharol144, 17 days ago

zoya had 2 1/2 kg of ladoo and manav had 3 4/5 kg of ladoo. each of them gave 1 1/2 kg of laddoos to bawana. how many kilograms of laddoos are still left with zoya and manav together?​

Answers

Answered by affanmobo
0

Answer:

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Class 12

>>Maths

>>Determinants

>>Applications of Matrices and Determinants

>>The cost of 4 kg onion, 3 kg wheat and 2

Question

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion,4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

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Solution

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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.

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