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The price of one kaya bun is RMx and the price of one butter bun is RMy. Suzi buys 6 kaya buns and 3 butter buns while Fauzi buys 4 kaya buns and 5 butter buns. Suzi pays RM8.40 and Fauzi pays RM1 less than Suzi. Using the matrix method, calculate the price of one kaya bun and the price of one butter bun.

Ask by Barber Barber. in Malaysia
Jan 23,2025

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Tutor-Verified Answer

Answer

One kaya bun costs RM 1.10 and one butter bun costs RM 0.60.

Solution

To determine the prices of one kaya bun (**RMx**) and one butter bun (**RMy**), we can set up a system of linear equations based on the purchases made by Suzi and Fauzi. We'll then solve this system using the matrix method. ### Step 1: Formulate the Equations From the problem: 1. **Suzi's Purchase:** \[ 6x + 3y = 8.40 \] 2. **Fauzi's Purchase:** \[ 4x + 5y = 7.40 \] *(Note: Fauzi pays RM1 less than Suzi, so \(8.40 - 1 = 7.40\))* ### Step 2: Represent as a Matrix Equation The system can be written in matrix form as: \[ \begin{bmatrix} 6 & 3 \\ 4 & 5 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 8.40 \\ 7.40 \\ \end{bmatrix} \] ### Step 3: Find the Inverse of the Coefficient Matrix Let \( A = \begin{bmatrix} 6 & 3 \\ 4 & 5 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} \). First, compute the determinant of \( A \): \[ \text{det}(A) = (6)(5) - (4)(3) = 30 - 12 = 18 \] The inverse of \( A \) is: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} = \frac{1}{18} \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \] ### Step 4: Multiply the Inverse by Vector \( \mathbf{b} \) \[ \begin{bmatrix} x \\ y \\ \end{bmatrix} = A^{-1} \mathbf{b} = \frac{1}{18} \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} \] Calculating the products: \[ x = \frac{1}{18} (5 \times 8.40 - 3 \times 7.40) = \frac{1}{18} (42.00 - 22.20) = \frac{19.80}{18} = 1.10 \] \[ y = \frac{1}{18} (-4 \times 8.40 + 6 \times 7.40) = \frac{1}{18} (-33.60 + 44.40) = \frac{10.80}{18} = 0.60 \] ### Conclusion - **Price of one kaya bun (x):** RM **1.10** - **Price of one butter bun (y):** RM **0.60** **Answer:** One kaya bun costs RM 1.10 and one butter bun costs RM 0.60.

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Bonus Knowledge

To solve this problem using matrices, we first need to set up the equations based on the purchases made by Suzi and Fauzi. From the information given: 1. Suzi's purchase: \[ 6x + 3y = 8.40 \] 2. Fauzi's purchase: \[ 4x + 5y = 8.40 - 1 = 7.40 \] Now we can write the system of equations: \[ \begin{align*} 6x + 3y &= 8.40 \quad (1) \\ 4x + 5y &= 7.40 \quad (2) \end{align*} \] Next, we can express this system in matrix form: \[ \begin{bmatrix} 6 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} \] Now we can solve for \(x\) (the price of one kaya bun) and \(y\) (the price of one butter bun) using the inverse of the coefficient matrix. First, calculate the determinant of the coefficient matrix: \[ \text{Det} = (6)(5) - (3)(4) = 30 - 12 = 18 \] Now, find the inverse of the coefficient matrix: \[ \begin{bmatrix} 6 & 3 \\ 4 & 5 \end{bmatrix}^{-1} = \frac{1}{18} \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \] Now we multiply this inverse matrix by the constants matrix: \[ \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} = \begin{bmatrix} (5)(8.40) + (-3)(7.40) \\ (-4)(8.40) + (6)(7.40) \end{bmatrix} \] Calculating these: 1. For \(x\): \[ 5(8.40) - 3(7.40) = 42 - 22.20 = 19.80 \] 2. For \(y\): \[ -4(8.40) + 6(7.40) = -33.60 + 44.40 = 10.80 \] Now, dividing by the determinant (18): \[ x = \frac{19.80}{18} = 1.10 \] \[ y = \frac{10.80}{18} = 0.60 \] Therefore, the price of one kaya bun is **RM1.10** and the price of one butter bun is **RM0.60**.

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