Two schools A and B want to awad their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award `₹ x` each , `₹y` each and `₹z` each for the three respective values to `3,2` and `1` students respectively with a total award money of `₹1600`. School B wants to spend `₹2300` to award its `4,1` and `3` students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is `₹900`, using matrices, find the award money for each value.

Answer

To solve the problem using matrices, we need to set up the equations based on the information provided. Let's break it down step by step.

### Step 1: Set Up the Equations
From the problem, we can derive the following equations based on the total awards given by each school:

1. For School A:
\[ 3x + 2y + z = 1600 \quad \text{(Equation 1)} \]

2. For School B:
\[ 4x + y + 3z = 2300 \quad \text{(Equation 2)} \]

3. The total award for one prize on each value:
\[ x + y + z = 900 \quad \text{(Equation 3)} \]

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

\[ A = \begin{pmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1600 \\ 2300 \\ 900 \end{pmatrix} \]

### Step 3: Find the Determinant of Matrix A
To find \(A^{-1}\), we first need to calculate the determinant of \(A\):

\[ \text{det}(A) = 3(1 \cdot 1 - 3 \cdot 1) - 2(4 \cdot 1 - 3 \cdot 1) + 1(4 \cdot 1 - 1 \cdot 3) \]
Calculating this step-by-step:

1. \(3(1 - 3) = 3(-2) = -6\)
2. \(-2(4 - 3) = -2(1) = -2\)
3. \(1(4 - 3) = 1(1) = 1\)

Thus,
\[ \text{det}(A) = -6 - 2 + 1 = -7 \]

### Step 4: Find the Adjoint of Matrix A
Next, we need to find the cofactor matrix of \(A\) and then take its transpose to get the adjoint.

The cofactor matrix \(C\) is calculated as follows:

\[ C = \begin{pmatrix} (1 \cdot 1 - 3 \cdot 1) & -(4 \cdot 1 - 3 \cdot 1) & (4 \cdot 1 - 1 \cdot 3) \\ -(2 \cdot 1 - 1 \cdot 1) & (3 \cdot 1 - 1 \cdot 1) & -(3 \cdot 1 - 2 \cdot 1) \\ (2 \cdot 3 - 1 \cdot 1) & -(3 \cdot 4 - 1 \cdot 2) & (3 \cdot 1 - 2 \cdot 4) \end{pmatrix} \]

Calculating each term:

1. \(C_{11} = 1 - 3 = -2\)
2. \(C_{12} = -(4 - 3) = -1\)
3. \(C_{13} = 4 - 3 = 1\)
4. \(C_{21} = -(2 - 1) = -1\)
5. \(C_{22} = 3 - 1 = 2\)
6. \(C_{23} = -(3 - 2) = -1\)
7. \(C_{31} = 6 - 1 = 5\)
8. \(C_{32} = -(12 - 2) = -10\)
9. \(C_{33} = 3 - 8 = -5\)

Thus, the cofactor matrix is:
\[ C = \begin{pmatrix} -2 & -1 & 1 \\ -1 & 2 & -1 \\ 5 & -10 & -5 \end{pmatrix} \]

Taking the transpose gives us the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} -2 & -1 & 5 \\ -1 & 2 & -10 \\ 1 & -1 & -5 \end{pmatrix} \]

### Step 5: Find the Inverse of Matrix A
Now, we can find \(A^{-1}\):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-7} \cdot \begin{pmatrix} -2 & -1 & 5 \\ -1 & 2 & -10 \\ 1 & -1 & -5 \end{pmatrix} \]

Thus,
\[ A^{-1} = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} & -\frac{5}{7} \\ \frac{1}{7} & -\frac{2}{7} & \frac{10}{7} \\ -\frac{1}{7} & \frac{1}{7} & \frac{5}{7} \end{pmatrix} \]

### Step 6: Multiply \(A^{-1}\) by B
Now we multiply \(A^{-1}\) by \(B\) to find \(X\):
\[ X = A^{-1}B \]

Calculating:
\[ X = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} & -\frac{5}{7} \\ \frac{1}{7} & -\frac{2}{7} & \frac{10}{7} \\ -\frac{1}{7} & \frac{1}{7} & \frac{5}{7} \end{pmatrix} \begin{pmatrix} 1600 \\ 2300 \\ 900 \end{pmatrix} \]

Calculating each component:
1. \(x = \frac{2}{7}(1600) + \frac{1}{7}(2300) - \frac{5}{7}(900) = \frac{3200 + 2300 - 4500}{7} = \frac{1000}{7} \approx 200\)
2. \(y = \frac{1}{7}(1600) - \frac{2}{7}(2300) + \frac{10}{7}(900) = \frac{1600 - 4600 + 9000}{7} = \frac{4000}{7} \approx 300\)
3. \(z = -\frac{1}{7}(1600) + \frac{1}{7}(2300) + \frac{5}{7}(900) = \frac{-1600 + 2300 + 4500}{7} = \frac{4200}{7} \approx 400\)

### Final Result
Thus, the award money for each value is:
- \(x = 200\)
- \(y = 300\)
- \(z = 400\)