Let's solve the problem **step-by-step**: --- ### **Given Data** Let: - \( x \) = operation hours of pitch A in a month - \( y \) = operation hours of pitch B in a month - \( z \) = operation hours of pitch C in a month Utility cost per hour: - Pitch A: RM 4 - Pitch B: RM 5 - Pitch C: RM 6 Other information: - Total utility cost in a month = RM 3450 - Operation hour of pitch A is **double** that of pitch B: \( x = 2y \) - Total operation hours: \( x + y + z = 700 \) --- ## **(a) Write in matrix form** ### **Formulating the equations** 1. **Total cost:** \[ 4x + 5y + 6z = 3450 \] 2. **A is double B:** \[ x = 2y \] 3. **Total hours:** \[ x + y + z = 700 \] --- ### **Convert to standard matrix form** Write as a system: \[ \begin{align*} 4x + 5y + 6z &= 3450 \\ x - 2y + 0z &= 0 \\ x + y + z &= 700 \\ \end{align*} \] Matrix form: \( \mathbf{A} \mathbf{X} = \mathbf{B} \) \[ \begin{bmatrix} 4 & 5 & 6 \\ 1 & -2 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 3450 \\ 0 \\ 700 \\ \end{bmatrix} \] --- ## **(b) Solve using Gauss-Jordan Elimination** Write as augmented matrix: \[ \left[ \begin{array}{ccc|c} 4 & 5 & 6 & 3450 \\ 1 & -2 & 0 & 0 \\ 1 & 1 & 1 & 700 \\ \end{array} \right] \] ### **Step 1: Make leading 1 in Row 1** Divide **Row 2** by 1 (already has leading 1): \[ \left[ \begin{array}{ccc|c} 4 & 5 & 6 & 3450 \\ 1 & -2 & 0 & 0 \\ 1 & 1 & 1 & 700 \\ \end{array} \right] \] ### **Step 2: Eliminate \( x \) from Row 1 and Row 3** - Row 1: \( R_1 - 4R_2 \) - Row 3: \( R_3 - R_2 \) Calculate: - \( R_1 - 4R_2: [4-4*1,\, 5-4*(-2),\, 6-4*0,\, 3450-4*0] \) - \( [0, 13, 6, 3450] \) - \( R_3 - R_2: [1-1,\, 1-(-2),\, 1-0,\, 700-0] \) - \( [0, 3, 1, 700] \) New matrix: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 0 & 0 \\ 0 & 13 & 6 & 3450 \\ 0 & 3 & 1 & 700 \\ \end{array} \right] \] ### **Step 3: Make leading 1 in Row 2** Divide Row 2 by 13: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 0 & 0 \\ 0 & 1 & \frac{6}{13} & \frac{3450}{13} \\ 0 & 3 & 1 & 700 \\ \end{array} \right] \] ### **Step 4: Eliminate \( y \) in Row 3** Row 3: \( R_3 - 3R_2 \) - \( [0, 3-3*1, 1-3*(6/13), 700-3*(3450/13)] \) - \( [0, 0, 1 - 18/13, 700 - 10350/13] \) - \( 1 - 18/13 = (13-18)/13 = -5/13 \) - \( 3 * 3450/13 = 10350/13 \) - \( 700 - 10350/13 = (700*13 - 10350)/13 = (9100 - 10350)/13 = -1250/13 \) So Row 3 becomes: \[ [0, 0, -5/13, -1250/13] \] ### **Step 5: Make leading 1 in Row 3** Multiply Row 3 by \(-13/5\): - \( [0, 0, 1, (-1250/13) * (-13/5)] = [0, 0, 1, 250] \) So, \( z = 250 \) --- ### **Step 6: Back Substitute to Row 2** From Row 2: \[ y + \frac{6}{13}z = \frac{3450}{13} \] Plug in \( z = 250 \): \[ y + \frac{6}{13} \times 250 = \frac{3450}{13} \] \[ y + \frac{1500}{13} = \frac{3450}{13} \] \[ y = \frac{3450}{13} - \frac{1500}{13} = \frac{1950}{13} = 150 \] --- ### **Step 7: Back Substitute to Row 1** From Row 1: \( x - 2y = 0 \implies x = 2y \) Since \( y = 150 \): \[ x = 2 \times 150 = 300 \] --- ## **Final Answers** \[ \boxed{ \begin{align*} x = 300 \text{ hours} \\ y = 150 \text{ hours} \\ z = 250 \text{ hours} \\ \end{align*} } \] --- ### **Summary Table** | Pitch | Operation Hours (per month) | |-------|----------------------------| | A | 300 | | B | 150 | | C | 250 |