Two schools A and B decided to award prizes to their students for three values honesty `(x)`, puntuality `(y)` and obedience `(z)`. School A decided to award a total of `₹11000` for the three values to `5,4` and `3` students respectively while school B decided to award `₹10700` for the three values to `4,3` and `5` students respectively. If all the three prizes together amount to `₹2700 `, then :
`(i)` Represent the above situation by a matrix equation and form linear equations, using matrix multiplication.
`(ii)` Is it possible to solve the system of equations so obtained, using matrix multiplication?

Answer

To solve the problem step by step, we will represent the situation using a matrix equation and form linear equations using matrix multiplication.

### Step 1: Define the Variables
Let:
- \( x \) = Prize for honesty
- \( y \) = Prize for punctuality
- \( z \) = Prize for obedience

### Step 2: Set Up the Matrix Equation
We can represent the situation in the form of a matrix equation \( AX = B \).

1. **Matrix A** (coefficient matrix):
\[ A = \begin{bmatrix} 5 & 4 & 3 \\ 4 & 3 & 5 \\ 1 & 1 & 1 \end{bmatrix} \]

2. **Matrix X** (variable matrix):
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

3. **Matrix B** (constant matrix):
\[ B = \begin{bmatrix} 11000 \\ 10700 \\ 2700 \end{bmatrix} \]

### Step 3: Form the Linear Equations
Now, we can write the equations by multiplying the matrices:

1. From the first row of \( A \):
\[ 5x + 4y + 3z = 11000 \]

2. From the second row of \( A \):
\[ 4x + 3y + 5z = 10700 \]

3. From the third row of \( A \):
\[ x + y + z = 2700 \]

Thus, we have the following system of linear equations:
1. \( 5x + 4y + 3z = 11000 \) (Equation 1)
2. \( 4x + 3y + 5z = 10700 \) (Equation 2)
3. \( x + y + z = 2700 \) (Equation 3)

### Step 4: Determine if the System Can Be Solved
To determine if the system of equations can be solved using matrix multiplication, we need to check if the coefficient matrix \( A \) is invertible.

Since we can express the system in the form \( AX = B \), and if \( A \) has a non-zero determinant, then the inverse \( A^{-1} \) exists, and we can find \( X \) as follows:
\[ X = A^{-1}B \]

### Conclusion
(i) The matrix equation and linear equations are formed as shown above.

(ii) Yes, it is possible to solve the system of equations using matrix multiplication, provided that the determinant of matrix \( A \) is non-zero.