The three commodities C1, C2 , C3 are purchased and sold by the three persons A, B and C. Mr. A purchases 4 units of C3 and sells 3 units of C1 and 5 units of C2 . Mr B purchases 3 units of C1 and sells 2 units of C2 and 1 units of C3. Mr C purchases 1 units of C2 and sells 4 units of C1 and 6 units of C3. In the purchase – sell process Mr. A suffered a loss of Rs 1,000; Mr B earns no profit and Mr C earns Rs 40,000. Find the prices per unit of these commodities by using matrix method.
Solution
Let,
Price of commodity C1 per unit be x
Price of commodity C2 per unit be y
Price of commodity C3 per unit be z
Amount asociated with sells and profit is considered positive while amount considered with purchase and loss is considered negative.
Equation for Mr.A
3x + 5y - 4z = -1000 ......(i)
Equation for Mr.B
-3x + 2y + z = 0 ......(ii)
Equation for Mr.C
4x - y + 6z = 40000 ......(iii)
Writing the equation in the matrix form
AX = B
A =
3
-3
4
5
2
-1
-4
1
6
X =
x
y
z
B =
-1000
0
40000
3
-3
4
5
2
-1
-4
1
6
x
y
z
=
-1000
0
40000
X = A⁻¹ B.
So, now we find A⁻¹.
2
-1
1
6
= 2 × 6 - 1 × -1 = 13
A₁₁ = (-1)¹⁺¹ M₁₁ = 13
M₁₂ =
-3
4
1
6
= -3 × 6 - 1 × 4 = -22
A₁₂ = (-1)¹⁺² M₁₂ = 22
M₁₃ =
-3
4
2
-1
= -3 × -1 - 2 × 4 = -5
A₁₃ = (-1)¹⁺³ M₁₃ = -5
M₂₁ =
5
-1
-4
6
= 5 × 6 - -4 × -1 = 26
A₂₁ = (-1)²⁺¹ M₂₁ = -26
M₂₂ =
3
4
-4
6
= 3 × 6 - -4 × 4 = 34
A₂₂ = (-1)²⁺² M₂₂ = 34
M₂₃ =
3
4
5
-1
= 3 × -1 - 5 × 4 = -23
A₂₃ = (-1)²⁺³ M₂₃ = 23
M₃₁ =
5
2
-4
1
= 5 × 1 - -4 × 2 = 13
A₃₁ = (-1)³⁺¹ M₃₁ = 13
M₃₂ =
3
-3
-4
1
= 3 × 1 - -4 × -3 = -9
A₃₂ = (-1)³⁺² M₃₂ = 9
M₃₃ =
3
-3
5
2
= 3 × 2 - 5 × -3 = 21
A₃₃ = (-1)³⁺³ M₃₃ = 21
The matrix of cofactors of A is CA =
13
-26
13
22
34
9
-5
23
21
As we know, "Adjoint of a matrix = transpose of its Cofactor Matrix."
∴ Adjoint of matrix A =
13
22
-5
-26
34
23
13
9
21
Now we find the determinant of the matrix A
|A| =
3
-3
4
5
2
-1
-4
1
6
= 3 × 2 × 6 + 5 × 1 × 4 + (-4) × (-3) × (-1) - 3 × 1 × (-1) - 5 × (-3) × 6 - (-4) × 2 × 4
= 36 + 20-12+3+90+32
= 169
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|
13
22
-5
-26
34
23
13
9
21
=
1
169
13
22
-5
-26
34
23
13
9
21
∴ Inverse matrix of A = A-¹ =
0.077
0.13
-0.03
-0.154
0.201
0.136
0.077
0.053
0.124
A-¹ × B =
0.077
0.13
-0.03
-0.154
0.201
0.136
0.077
0.053
0.124
×
-1000
0
40000
A-¹ × B =
0.077×(-1000)+(-0.154)×0+0.077×40000
0.13×(-1000)+0.201×0+0.053×40000
(-0.03)×(-1000)+0.136×0+0.124×40000
A-¹ × B =
-77+0+3080
-130+0+2120
30+0+4960
∴ A-¹ B =
3003
1990
4990
∴x = 3003
∴y = 1990
∴z = 4990
∴y = 1990
∴z = 4990
Hence the price of C1 per unit = 3003
The price of C2 per unit = 1990
The price of C3 per unit = 4990
The price of C2 per unit = 1990
The price of C3 per unit = 4990
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