Three different chemiclas are mixed in a specific proportion to produce 100 units of a product. The company has to meet a constraint in the production of the product that chemical X should be one-third of the difference between chemicals Y and Z used. The cost of chemicals X, Y and Z are respecvtively Rs 50 Rs 20 and Rs 40 and the total cost of all the chemicals used is Rs 2900. Determine each unit of chemical X, Y and Z used by matrix method.
Solution
Let,
x units of X
y units of Y and
z units of Z are used to produce 100 units of the product.
So, x + y + z = 100 ......(i)
Second condition is :
Chemical X should be one-third of the difference between chemicals Y and Z used
x =
1
3(y - z)
Or, 3x = y-z
So, 3x - y + z = 0 ......(ii)
Third condition is :
Total cost = 2900 which is sum of cost of X, Y and Z
So, 50x + 20y + 40z = 2900 ......(iii)
Writing the equation in the matrix form
AX = B
A =
1
3
50
1
-1
20
1
1
40
X =
x
y
z
B =
100
0
2900
1
3
50
1
-1
20
1
1
40
x
y
z
=
100
0
2900
X = A⁻¹ B.
So, now we find A⁻¹.
-1
20
1
40
= -1 × 40 - 1 × 20 = -60
A₁₁ = (-1)¹⁺¹ M₁₁ = -60
M₁₂ =
3
50
1
40
= 3 × 40 - 1 × 50 = 70
A₁₂ = (-1)¹⁺² M₁₂ = -70
M₁₃ =
3
50
-1
20
= 3 × 20 - -1 × 50 = 110
A₁₃ = (-1)¹⁺³ M₁₃ = 110
M₂₁ =
1
20
1
40
= 1 × 40 - 1 × 20 = 20
A₂₁ = (-1)²⁺¹ M₂₁ = -20
M₂₂ =
1
50
1
40
= 1 × 40 - 1 × 50 = -10
A₂₂ = (-1)²⁺² M₂₂ = -10
M₂₃ =
1
50
1
20
= 1 × 20 - 1 × 50 = -30
A₂₃ = (-1)²⁺³ M₂₃ = 30
M₃₁ =
1
-1
1
1
= 1 × 1 - 1 × -1 = 2
A₃₁ = (-1)³⁺¹ M₃₁ = 2
M₃₂ =
1
3
1
1
= 1 × 1 - 1 × 3 = -2
A₃₂ = (-1)³⁺² M₃₂ = 2
M₃₃ =
1
3
1
-1
= 1 × -1 - 1 × 3 = -4
A₃₃ = (-1)³⁺³ M₃₃ = -4
The matrix of cofactors of A is CA =
-60
-20
2
-70
-10
2
110
30
-4
As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."
∴ Adjoint of matrix A =
-60
-70
110
-20
-10
30
2
2
-4
Now we find the determinant of the matrix A
|A| =
1
3
50
1
-1
20
1
1
40
= 1 × (-1) × 40 + 1 × 1 × 50 + 1 × 3 × 20 - 1 × 1 × 20 - 1 × 3 × 40 - 1 × (-1) × 50
= -40 + 50 + 60-20-120+50
= -20
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|
-60
-70
110
-20
-10
30
2
2
-4
=
1
-20
-60
-70
110
-20
-10
30
2
2
-4
∴ Inverse matrix of A = A-¹ =
3
3.5
-5.5
1
0.5
-1.5
-0.1
-0.1
0.2
A-¹ × B =
3
3.5
-5.5
1
0.5
-1.5
-0.1
-0.1
0.2
×
100
0
2900
A-¹ × B =
3×100+1×0+(-0.1)×2900
3.5×100+0.5×0+(-0.1)×2900
(-5.5)×100+(-1.5)×0+0.2×2900
A-¹ × B =
300+0-290
350+0-290
-550+0+580
∴ A-¹ B =
10
60
30
∴x = 10
∴y = 60
∴z = 30
∴y = 60
∴z = 30
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