Diary Corporation is planning to produce 1200 kgs of baby food mixing 3 nutrients A, B and C. The 3 nutrient components per kg cost Rs 12, Rs 16 and Rs 14 respectively. The final output of 1 kg of baby food pack must contain nutrient C to be twice that of nutrient A . The budget available to buy component nutrients are to the corporation is Rs 18,480. Find amount of each nutrient which should be mixed in the final output by using matrix method.
Solution
Let,
a kgs of A
b kgs of B and
c kgs of C are used to produce 1200 kgs of the baby food.
So, a + b + c = 1200 ......(i)
Second condition is :
Total cost of the baby food is Rs. 18,480
So, 12a + 16b + 14c = 18480 ......(ii)
Third condition is :
Nutrient C should be used twice nutrient A.
So, c = 2a
2a - c = 0 ......(iii)
Writing the equation in the matrix form
AX = B
A =
1
12
2
1
16
0
1
14
-1
X =
a
b
c
B =
1200
18480
0
1
12
2
1
16
0
1
14
-1
a
b
c
=
1200
18480
0
X = A⁻¹ B.
So, now we find A⁻¹.
16
0
14
-1
= 16 × -1 - 14 × 0 = -16
A₁₁ = (-1)¹⁺¹ M₁₁ = -16
M₁₂ =
12
2
14
-1
= 12 × -1 - 14 × 2 = -40
A₁₂ = (-1)¹⁺² M₁₂ = 40
M₁₃ =
12
2
16
0
= 12 × 0 - 16 × 2 = -32
A₁₃ = (-1)¹⁺³ M₁₃ = -32
M₂₁ =
1
0
1
-1
= 1 × -1 - 1 × 0 = -1
A₂₁ = (-1)²⁺¹ M₂₁ = 1
M₂₂ =
1
2
1
-1
= 1 × -1 - 1 × 2 = -3
A₂₂ = (-1)²⁺² M₂₂ = -3
M₂₃ =
1
2
1
0
= 1 × 0 - 1 × 2 = -2
A₂₃ = (-1)²⁺³ M₂₃ = 2
M₃₁ =
1
16
1
14
= 1 × 14 - 1 × 16 = -2
A₃₁ = (-1)³⁺¹ M₃₁ = -2
M₃₂ =
1
12
1
14
= 1 × 14 - 1 × 12 = 2
A₃₂ = (-1)³⁺² M₃₂ = -2
M₃₃ =
1
12
1
16
= 1 × 16 - 1 × 12 = 4
A₃₃ = (-1)³⁺³ M₃₃ = 4
The matrix of cofactors of A is CA =
-16
1
-2
40
-3
-2
-32
2
4
As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."
∴ Adjoint of matrix A =
-16
40
-32
1
-3
2
-2
-2
4
Now we find the determinant of the matrix A
|A| =
1
12
2
1
16
0
1
14
-1
= 1 × 16 × (-1) + 1 × 14 × 2 + 1 × 12 × 0 - 1 × 14 × 0 - 1 × 12 × (-1) - 1 × 16 × 2
= -16 + 28 + 0-0+12-32
= -8
As determinant of the matrix ≠ 0. Hence its inverse exists.
Inverse of a matrix is scalar multiplication of its adjugate matrix with the reciprocal of its determinant.
Inverse of matrix A = A-¹
=
1
|A|
-16
40
-32
1
-3
2
-2
-2
4
=
1
-8
-16
40
-32
1
-3
2
-2
-2
4
∴ Inverse matrix of A = A-¹ =
2
-5
4
-0.125
0.375
-0.25
0.25
0.25
-0.5
A-¹ × B =
2
-5
4
-0.125
0.375
-0.25
0.25
0.25
-0.5
×
1200
18480
0
A-¹ × B =
2×1200+(-0.125)×18480+0.25×0
(-5)×1200+0.375×18480+0.25×0
4×1200+(-0.25)×18480+(-0.5)×0
A-¹ × B =
2400-2310+0
-6000+6930+0
4800-4620+0
∴ A-¹ B =
90
930
180
∴a = 90
∴b = 930
∴c = 180
∴b = 930
∴c = 180
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