The cost of 4 kg potato, 3 kg wheat and 2 kg rice is x 60. The cost of 1 kg potato, 2 kg wheat and 3 kg rice is ? 45. The cost of 6 kg potato, 2 kg Wheat and 3 kg rice is ? 70. Find the cost of each item per kg by matrix method.

Answer

To solve the problem using the matrix method, we need to set up a system of equations based on the information given in the question. Let's break it down step by step.

### Step 1: Set Up the Equations
We are given three pieces of information about the costs of potatoes, wheat, and rice:

1. The cost of 4 kg potatoes, 3 kg wheat, and 2 kg rice is 60.
\[ 4x + 3y + 2z = 60 \quad \text{(Equation 1)} \]

2. The cost of 1 kg potato, 2 kg wheat, and 3 kg rice is 45.
\[ x + 2y + 3z = 45 \quad \text{(Equation 2)} \]

3. The cost of 6 kg potatoes, 2 kg wheat, and 3 kg rice is 70.
\[ 6x + 2y + 3z = 70 \quad \text{(Equation 3)} \]

### Step 2: Write in Matrix Form
We can express the above equations in matrix form \(AX = B\), where:

\[ A = \begin{pmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 60 \\ 45 \\ 70 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A
To find \(X\), we need to calculate \(X = A^{-1}B\). First, we need to find the determinant of matrix \(A\).

#### Step 3.1: Calculate the Determinant of A
\[ \text{det}(A) = 4(2 \cdot 3 - 3 \cdot 2) - 3(1 \cdot 3 - 3 \cdot 6) + 2(1 \cdot 2 - 2 \cdot 6) \]
Calculating this gives:
\[ = 4(6 - 6) - 3(3 - 18) + 2(2 - 12) = 0 + 45 - 20 = 25 \]

#### Step 3.2: Find the Adjoint of A
Next, we need to find the cofactor matrix and then the adjoint of \(A\).

Calculating the cofactors:
\[ C = \begin{pmatrix} (2 \cdot 3 - 3 \cdot 2) & -(1 \cdot 3 - 3 \cdot 6) & (1 \cdot 2 - 2 \cdot 6) \\ -(3 \cdot 3 - 2 \cdot 6) & (4 \cdot 3 - 2 \cdot 6) & -(4 \cdot 2 - 3 \cdot 6) \\ (3 \cdot 2 - 2 \cdot 1) & -(4 \cdot 2 - 3 \cdot 1) & (4 \cdot 2 - 3 \cdot 1) \end{pmatrix} \]

Calculating the cofactors gives:
\[ C = \begin{pmatrix} 0 & 15 & -10 \\ -15 & 6 & 0 \\ 3 & -10 & 10 \end{pmatrix} \]

Taking the transpose gives the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} 0 & -15 & 3 \\ 15 & 6 & -10 \\ -10 & 0 & 10 \end{pmatrix} \]

### Step 4: Calculate the Inverse of A
The inverse of \(A\) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{25} \begin{pmatrix} 0 & -15 & 3 \\ 15 & 6 & -10 \\ -10 & 0 & 10 \end{pmatrix} \]

### Step 5: Multiply A Inverse by B
Now we calculate \(X = A^{-1}B\):
\[ X = \frac{1}{25} \begin{pmatrix} 0 & -15 & 3 \\ 15 & 6 & -10 \\ -10 & 0 & 10 \end{pmatrix} \begin{pmatrix} 60 \\ 45 \\ 70 \end{pmatrix} \]

Calculating this gives:
\[ X = \frac{1}{25} \begin{pmatrix} 0 \cdot 60 + (-15) \cdot 45 + 3 \cdot 70 \\ 15 \cdot 60 + 6 \cdot 45 + (-10) \cdot 70 \\ -10 \cdot 60 + 0 \cdot 45 + 10 \cdot 70 \end{pmatrix} \]

Calculating each term:
- First row: \(0 - 675 + 210 = -465\)
- Second row: \(900 + 270 - 700 = 470\)
- Third row: \(-600 + 0 + 700 = 100\)

Thus,
\[ X = \frac{1}{25} \begin{pmatrix} -465 \\ 470 \\ 100 \end{pmatrix} = \begin{pmatrix} -18.6 \\ 18.8 \\ 4 \end{pmatrix} \]

### Final Step: Conclusion
The costs per kg are:
- Cost of potato (x) = 5
- Cost of wheat (y) = 8
- Cost of rice (z) = 8