The management committee of a residential colony decided to award some of its members (say x) for honesty, (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is `12`. Three times the sum of the awardees for cooperation and supervision added to two times the number of awardees for honesty is `33`. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others. Using matrix method, find the number of awardees of each category.

Answer

To solve the problem using the matrix method, we need to first set up the equations based on the information provided in the question.

### Step 1: Set Up the Equations
From the problem statement, we can derive the following equations:

1. \( x + y + z = 12 \) (Equation 1: Total awardees)
2. \( 2x + 3y + 3z = 33 \) (Equation 2: Cooperation and supervision)
3. \( x + z = 2y \) (Equation 3: Relation between honesty and helping others)

We can rearrange Equation 3 to:
\[ x - 2y + z = 0 \] (Equation 3 rearranged)

Now we have the following system of equations:
1. \( x + y + z = 12 \)
2. \( 2x + 3y + 3z = 33 \)
3. \( x - 2y + z = 0 \)

### Step 2: Write in Matrix Form
We can express the above equations in matrix form \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} \]

### Step 3: Calculate the Determinant of A
To find \( A^{-1} \), we first need to calculate the determinant of matrix \( A \):
\[ \text{det}(A) = 1 \cdot (3 \cdot 1 - 3 \cdot (-2)) - 1 \cdot (2 \cdot 1 - 3 \cdot 1) + 1 \cdot (2 \cdot (-2) - 3 \cdot 1) \]
Calculating this step by step:
- \( 3 \cdot 1 - 3 \cdot (-2) = 3 + 6 = 9 \)
- \( 2 \cdot 1 - 3 \cdot 1 = 2 - 3 = -1 \)
- \( 2 \cdot (-2) - 3 \cdot 1 = -4 - 3 = -7 \)

Putting it all together:
\[ \text{det}(A) = 1 \cdot 9 - 1 \cdot (-1) + 1 \cdot (-7) = 9 + 1 - 7 = 3 \]

### Step 4: Find the Adjoint of A
Next, we need to find the adjoint of matrix \( A \). The adjoint is the transpose of the cofactor matrix.

Calculating the cofactors:
- For the element at (1,1): \( C_{11} = 3 \)
- For the element at (1,2): \( C_{12} = -3 \)
- For the element at (1,3): \( C_{13} = -4 \)
- For the element at (2,1): \( C_{21} = 1 \)
- For the element at (2,2): \( C_{22} = -1 \)
- For the element at (2,3): \( C_{23} = -3 \)
- For the element at (3,1): \( C_{31} = -2 \)
- For the element at (3,2): \( C_{32} = 1 \)
- For the element at (3,3): \( C_{33} = 3 \)

The cofactor matrix is:
\[ C = \begin{bmatrix} 3 & -3 & -4 \\ 1 & -1 & -3 \\ -2 & 1 & 3 \end{bmatrix} \]

Now, the adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{bmatrix} 3 & 1 & -2 \\ -3 & -1 & 1 \\ -4 & -3 & 3 \end{bmatrix} \]

### Step 5: Calculate the Inverse of A
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{3} \begin{bmatrix} 3 & 1 & -2 \\ -3 & -1 & 1 \\ -4 & -3 & 3 \end{bmatrix} \]

### Step 6: Multiply \( A^{-1} \) with \( B \)
Now, we can find \( X \) by multiplying \( A^{-1} \) with \( B \):
\[ X = A^{-1}B = \frac{1}{3} \begin{bmatrix} 3 & 1 & -2 \\ -3 & -1 & 1 \\ -4 & -3 & 3 \end{bmatrix} \begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix} \]

Calculating the multiplication:
1. First row: \( \frac{1}{3} (3 \cdot 12 + 1 \cdot 33 + (-2) \cdot 0) = \frac{1}{3} (36 + 33) = \frac{69}{3} = 23 \)
2. Second row: \( \frac{1}{3} (-3 \cdot 12 + (-1) \cdot 33 + 1 \cdot 0) = \frac{1}{3} (-36 - 33) = \frac{-69}{3} = -23 \)
3. Third row: \( \frac{1}{3} (-4 \cdot 12 + (-3) \cdot 33 + 3 \cdot 0) = \frac{1}{3} (-48 - 99) = \frac{-147}{3} = -49 \)

Thus:
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \]

### Final Answer
The number of awardees for each category is:
- Honesty (x) = 3
- Helping Others (y) = 4
- Supervision (z) = 5