If the total cost of 6 kg potatoes and 4 kg of tomatoes is Rs. 200 and the total cost of 1 kg
of potatoes and 1 kg tomatoes is 42.
Find the price of per kg of each vegetable by using
determinant.

Verified Answer :

Alright! Let's solve the given system of equations step-by-step using determinants.

We have two equations based on the given information:

1. \(\( 6x + 4y = 200 \)\) (Equation 1)
2. \(\( x + y = 42 \)\) (Equation 2)

where:
- \(\( x \)\) represents the price per kg of potatoes.
- \(\( y \)\) represents the price per kg of tomatoes.

We will solve these equations using determinants, following Cramer's rule.

### Step 1: Write the system of equations in matrix form

The system of equations can be written in matrix form as:
\(\[ \begin{bmatrix} 6 & 4 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 200 \\ 42 \end{bmatrix} \]\)

### Step 2: Find the determinant of the coefficient matrix

The coefficient matrix is:
\(\[ A = \begin{bmatrix} 6 & 4 \\ 1 & 1 \end{bmatrix} \]\)

The determinant of \(\( A \)\) (\(\( \text{det}(A) \)\)) is calculated as:
\(\[ \text{det}(A) = (6 \cdot 1) - (4 \cdot 1) = 6 - 4 = 2 \]\)

### Step 3: Find the determinants of matrices \(\( A_x \)\) and \(\( A_y \)\)

To find the determinants of \(\( A_x \)\) and \(\( A_y \)\), we replace the appropriate columns of \(\( A \)\) with the constants from the right-hand side of the equations.

#### Matrix \(\( A_x \)\)
\(\[ A_x = \begin{bmatrix} 200 & 4 \\ 42 & 1 \end{bmatrix} \]\)

The determinant of \(\( A_x \)\) (\(\( \text{det}(A_x) \)\)) is:
\(\[ \text{det}(A_x) = (200 \cdot 1) - (4 \cdot 42) = 200 - 168 = 32 \]\)

#### Matrix \(\( A_y \)\)
\(\[ A_y = \begin{bmatrix} 6 & 200 \\ 1 & 42 \end{bmatrix} \]\)

The determinant of \(\( A_y \)\) (\(\( \text{det}(A_y) \)\)) is:
\(\[ \text{det}(A_y) = (6 \cdot 42) - (200 \cdot 1) = 252 - 200 = 52 \]\)

### Step 4: Apply Cramer's Rule

Using Cramer's Rule, we can find \(\( x \)\) and \(\( y \)\) by:
\(\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{32}{2} = 16 \]\)
\(\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{52}{2} = 26 \]\)

Thus, the price per kg of potatoes \(\( (x) \)\) is Rs. 16, and the price per kg of tomatoes \(\( (y) \)\) is Rs. 26.

### Final Answer

- The price of 1 kg of potatoes is Rs. 16.
- The price of 1 kg of tomatoes is Rs. 26.