A subjective test consists of two sections, A and B. Each question in Section A and Section B are allotted $$ x $$ marks and $$ y $$ marks respectively. Ali answered correctly 8 questions from Section A and 7 questions from Section B. Aiman answered correctly 9 questions from Section A and 5 questions from Section B. Ali and Aiman scored 59 marks and 52 marks respectively. Using the matrix method, calculate the values of $$ x $$ and $$ y $$.

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To determine the values of $$ x $$ and $$ y $$ using the **matrix method**, we'll follow these steps:

1. **Set Up the Equations**

From the problem statement, we can form the following system of linear equations based on the scores of Ali and Aiman:

$$
   \begin{cases}
   8x + 7y = 59 & \text{(Ali's score)} \
   9x + 5y = 52 & \text{(Aiman's score)}
   \end{cases}
   $$

2. **Represent the System in Matrix Form**

The system can be written as:

$$
   \begin{bmatrix}
   8 & 7 \
   9 & 5 \
   \end{bmatrix}
   \begin{bmatrix}
   x \
   y \
   \end{bmatrix}
   =
   \begin{bmatrix}
   59 \
   52 \
   \end{bmatrix}
   $$

Let's denote:
   
   $$
   \mathbf{A} = \begin{bmatrix} 8 & 7 \ 9 & 5 \end{bmatrix}, \quad 
   \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix}, \quad 
   \mathbf{B} = \begin{bmatrix} 59 \ 52 \end{bmatrix}
   $$
   
   So, the matrix equation is:
   
   $$
   \mathbf{A} \mathbf{X} = \mathbf{B}
   $$

3. **Find the Inverse of Matrix $$ \mathbf{A} $$**

First, compute the determinant of $$ \mathbf{A} $$:

$$
   \text{det}(\mathbf{A}) = (8)(5) - (9)(7) = 40 - 63 = -23
   $$

Since the determinant is not zero, $$ \mathbf{A} $$ is invertible. The inverse of $$ \mathbf{A} $$ is:

$$
   \mathbf{A}^{-1} = \frac{1}{\text{det}(\mathbf{A})}
   \begin{bmatrix}
   5 & -7 \
   -9 & 8 \
   \end{bmatrix}
   =
   \frac{1}{-23}
   \begin{bmatrix}
   5 & -7 \
   -9 & 8 \
   \end{bmatrix}
   $$

4. **Multiply Both Sides by $$ \mathbf{A}^{-1} $$**

To solve for $$ \mathbf{X} $$:

$$
   \mathbf{X} = \mathbf{A}^{-1} \mathbf{B}
   $$

Let's compute this multiplication:

$$
   \mathbf{X} = \frac{1}{-23}
   \begin{bmatrix}
   5 & -7 \
   -9 & 8 \
   \end{bmatrix}
   \begin{bmatrix}
   59 \
   52 \
   \end{bmatrix}
   $$

5. **Calculate the Products**

- **For $$ x $$:**

$$
     x = \frac{1}{-23} \left(5 \times 59 + (-7) \times 52\right) = \frac{1}{-23} \left(295 - 364\right) = \frac{1}{-23} \times (-69) = 3
     $$

- **For $$ y $$:**

$$
     y = \frac{1}{-23} \left((-9) \times 59 + 8 \times 52\right) = \frac{1}{-23} \left(-531 + 416\right) = \frac{1}{-23} \times (-115) = 5
     $$

6. **Solution**

$$
   x = 3 \quad \text{and} \quad y = 5
   $$

**Therefore, each question in Section A is worth 3 marks, and each question in Section B is worth 5 marks.**
Each question in Section A is worth 3 marks, and each question in Section B is worth 5 marks.

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