To solve the given problem, we will represent the situation using a system of equations and then use determinants to find the cost of each item per kg.
### Step 1: Define Variables
Let:
- \( x \) = cost of 1 kg of sugar (in ₹)
- \( y \) = cost of 1 kg of wheat (in ₹)
- \( z \) = cost of 1 kg of rice (in ₹)
### Step 2: Set Up Equations Based on the Problem
From the problem statement, we can form the following equations:
1. From the first statement: "₹ 240 is the cost of 5 kg sugar, 5 kg wheat, and 2 kg rice."
\[
5x + 5y + 2z = 240 \quad \text{(Equation 1)}
\]
2. From the second statement: "The cost of 4 kg rice, 2 kg sugar, and 5 kg wheat is ₹ 190."
\[
4z + 2x + 5y = 190 \quad \text{(Equation 2)}
\]
3. From the third statement: "The cost of 3 kg wheat, 2 kg rice, and 4 kg sugar is ₹ 190."
\[
4x + 3y + 2z = 190 \quad \text{(Equation 3)}
\]
### Step 3: Write the System of Equations in Matrix Form
We can represent the system of equations in the form \( AX = B \), where:
\[
A = \begin{bmatrix}
5 & 5 & 2 \\
2 & 5 & 4 \\
4 & 3 & 2
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}, \quad
B = \begin{bmatrix}
240 \\
190 \\
190
\end{bmatrix}
\]
### Step 4: Calculate the Determinant of Matrix A
To find the values of \( x, y, z \), we first need to calculate the determinant of matrix \( A \).
\[
\text{det}(A) = 5 \begin{vmatrix}
5 & 4 \\
3 & 2
\end{vmatrix} - 5 \begin{vmatrix}
2 & 4 \\
4 & 2
\end{vmatrix} + 2 \begin{vmatrix}
2 & 5 \\
4 & 3
\end{vmatrix}
\]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 5 & 4 \\ 3 & 2 \end{vmatrix} = (5 \cdot 2) - (4 \cdot 3) = 10 - 12 = -2 \)
2. \( \begin{vmatrix} 2 & 4 \\ 4 & 2 \end{vmatrix} = (2 \cdot 2) - (4 \cdot 4) = 4 - 16 = -12 \)
3. \( \begin{vmatrix} 2 & 5 \\ 4 & 3 \end{vmatrix} = (2 \cdot 3) - (5 \cdot 4) = 6 - 20 = -14 \)
Now substituting back:
\[
\text{det}(A) = 5(-2) - 5(-12) + 2(-14) = -10 + 60 - 28 = 22
\]
### Step 5: Find the Adjoint of Matrix A
Next, we need to find the adjoint of matrix \( A \). The adjoint is found using the cofactor matrix.
Calculating the cofactors:
- \( C_{11} = -2 \)
- \( C_{12} = 12 \)
- \( C_{13} = -14 \)
- \( C_{21} = -4 \)
- \( C_{22} = 22 \)
- \( C_{23} = 5 \)
- \( C_{31} = 10 \)
- \( C_{32} = -16 \)
- \( C_{33} = 15 \)
Thus, the adjoint matrix is:
\[
\text{adj}(A) = \begin{bmatrix}
-2 & 12 & -14 \\
-4 & 22 & 5 \\
10 & -16 & 15
\end{bmatrix}
\]
### Step 6: Calculate the Inverse of Matrix A
The inverse of \( A \) is given by:
\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{22} \begin{bmatrix}
-2 & 12 & -14 \\
-4 & 22 & 5 \\
10 & -16 & 15
\end{bmatrix}
\]
### Step 7: Multiply \( A^{-1} \) by \( B \)
Now we will find \( X \) by multiplying \( A^{-1} \) with \( B \):
\[
X = A^{-1}B = \frac{1}{22} \begin{bmatrix}
-2 & 12 & -14 \\
-4 & 22 & 5 \\
10 & -16 & 15
\end{bmatrix} \begin{bmatrix}
240 \\
190 \\
190
\end{bmatrix}
\]
Calculating the product:
1. First row: \( -2(240) + 12(190) - 14(190) = -480 + 2280 - 2660 = -860 \)
2. Second row: \( -4(240) + 22(190) + 5(190) = -960 + 4180 + 950 = 3170 \)
3. Third row: \( 10(240) - 16(190) + 15(190) = 2400 - 3040 + 2850 = 2210 \)
Thus,
\[
X = \frac{1}{22} \begin{bmatrix}
-860 \\
3170 \\
2210
\end{bmatrix}
\]
### Step 8: Simplify to Find \( x, y, z \)
Calculating the values:
\[
x = \frac{-860}{22}, \quad y = \frac{3170}{22}, \quad z = \frac{2210}{22}
\]
Calculating:
- \( x = 30 \) (cost of 1 kg sugar)
- \( y = 10 \) (cost of 1 kg wheat)
- \( z = 20 \) (cost of 1 kg rice)
### Final Answer
The costs per kg are:
- Cost of 1 kg of sugar = ₹ 30
- Cost of 1 kg of wheat = ₹ 10
- Cost of 1 kg of rice = ₹ 20
