Find the cost of sugar and wheat per kg, if the cost of `7kg`.of sugar and `3kg`.of wheat is `₹ 240` and cost of `7kg` of wheat and `3kg`.of sugar is `₹ 160`. (Use determinants)

Answer

To solve the problem of finding the cost of sugar and wheat per kg using determinants, we can follow these steps:

### Step 1: Set Up the Equations
Let the cost of 1 kg of sugar be \( x \) (in ₹) and the cost of 1 kg of wheat be \( y \) (in ₹). We can form the following equations based on the information given:

1. For 7 kg of sugar and 3 kg of wheat costing ₹ 240:
\[ 7x + 3y = 240 \quad \text{(Equation 1)} \]

2. For 7 kg of wheat and 3 kg of sugar costing ₹ 160:
\[ 3x + 7y = 160 \quad \text{(Equation 2)} \]

### Step 2: Write in Matrix Form
We can express the system of equations in matrix form as follows:
\[ \begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 240 \\ 160 \end{bmatrix} \]

### Step 3: Calculate the Determinant
The determinant \( D \) of the coefficient matrix is calculated as:
\[ D = \begin{vmatrix} 7 & 3 \\ 3 & 7 \end{vmatrix} = (7 \cdot 7) - (3 \cdot 3) = 49 - 9 = 40 \]

### Step 4: Calculate the Adjoint
The adjoint of the matrix is given by swapping the diagonal elements and changing the signs of the off-diagonal elements:
\[ \text{Adjoint} = \begin{bmatrix} 7 & -3 \\ -3 & 7 \end{bmatrix} \]

### Step 5: Calculate the Inverse
The inverse of the matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{D} \cdot \text{Adjoint} = \frac{1}{40} \begin{bmatrix} 7 & -3 \\ -3 & 7 \end{bmatrix} \]

### Step 6: Multiply the Inverse by the Constant Matrix
Now, we will multiply the inverse matrix by the constant matrix:
\[ \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} 240 \\ 160 \end{bmatrix} \]
Calculating this gives:
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 7 & -3 \\ -3 & 7 \end{bmatrix} \begin{bmatrix} 240 \\ 160 \end{bmatrix} \]

Calculating the multiplication:
- For \( x \):
\[ x = \frac{1}{40} (7 \cdot 240 - 3 \cdot 160) = \frac{1}{40} (1680 - 480) = \frac{1200}{40} = 30 \]

- For \( y \):
\[ y = \frac{1}{40} (-3 \cdot 240 + 7 \cdot 160) = \frac{1}{40} (-720 + 1120) = \frac{400}{40} = 10 \]

### Final Values
Thus, the cost of 1 kg of sugar is \( ₹ 30 \) and the cost of 1 kg of wheat is \( ₹ 10 \).