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Sagot :
Sure, let's tackle the problem step by step.
### PART A
#### Copy and complete the table
| | Mrs Ndoro | Mrs Kori | Miss Ndlovu |
|-------------------|-----------|----------|-------------|
| Bananas (kg) | 1.2 | 2.2 | 1.8 |
| Apples (kg) | 0.8 | 2.0 | 1.6 |
| Total Paid (\[tex]$) | 2.40 | 5.20 | 4.2 | The missing value for the total amount paid by Miss Ndlovu is \(4.2\). #### Define the variables Let: - \( x \) be the price per kilogram for Bananas. - \( y \) be the price per kilogram for Apples. #### Form separate equations using the information from each column Mrs Ndoro's Equation: \[ 1.2x + 0.8y = 2.40 \] Mrs Kori's Equation: \[ 2.2x + 2.0y = 5.20 \] Miss Ndlovu's Equation: \[ 1.8x + 1.6y = 4.2 \] ### PART B #### Solve for the unknown variables using the matrix method ##### Pulling out the matrix To solve the system of linear equations defined by Mrs Ndoro and Mrs Kori, we can write it in matrix form \( A \mathbf{x} = \mathbf{b} \). \[ A = \begin{pmatrix} 1.2 & 0.8 \\ 2.2 & 2.0 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} \] \[ \mathbf{b} = \begin{pmatrix} 2.40 \\ 5.20 \end{pmatrix} \] ##### Finding the determinant of the drawn matrix To find \( x \) and \( y \), we first need to check that the determinant of \( A \) is non-zero. \[ \text{Det}(A) = \begin{vmatrix} 1.2 & 0.8 \\ 2.2 & 2.0 \end{vmatrix} = (1.2 \cdot 2.0) - (0.8 \cdot 2.2) \] \[ \text{Det}(A) = 2.4 - 1.76 = 0.64 \] Since the determinant is \( 0.64 \), which is not zero, we can proceed to solve the system using the inverse of \( A \). #### Solving the system The inverse of matrix \( A \) is given by \[ A^{-1} = \frac{1}{\text{Det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] where \( a = 1.2 \), \( b = 0.8 \), \( c = 2.2 \), and \( d = 2.0 \). \[ A^{-1} = \frac{1}{0.64} \begin{pmatrix} 2.0 & -0.8 \\ -2.2 & 1.2 \end{pmatrix} = \begin{pmatrix} 3.125 & -1.25 \\ -3.4375 & 1.875 \end{pmatrix} \] Now we can find \( \mathbf{x} \) by multiplying \( A^{-1} \) with \( \mathbf{b} \): \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} 3.125 & -1.25 \\ -3.4375 & 1.875 \end{pmatrix} \begin{pmatrix} 2.40 \\ 5.20 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} (3.125 \cdot 2.40) + (-1.25 \cdot 5.20) \\ (-3.4375 \cdot 2.40) + (1.875 \cdot 5.20) \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} 7.5 - 6.5 \\ -8.25 + 9.75 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} 1.00 \\ 1.50 \end{pmatrix} \] Thus, the price per kilogram for Bananas is \( \$[/tex]1.00 \) and the price per kilogram for Apples is [tex]\( \$1.50 \)[/tex].
Finally, we use these prices to calculate the total amount paid by Miss Ndlovu:
[tex]\[ 1.8x + 1.6y = 1.8 \cdot 1.00 + 1.6 \cdot 1.50 = 1.8 + 2.4 = 4.2 \][/tex]
So, Miss Ndlovu paid [tex]\( \$4.20 \)[/tex] for her fruits.
### PART A
#### Copy and complete the table
| | Mrs Ndoro | Mrs Kori | Miss Ndlovu |
|-------------------|-----------|----------|-------------|
| Bananas (kg) | 1.2 | 2.2 | 1.8 |
| Apples (kg) | 0.8 | 2.0 | 1.6 |
| Total Paid (\[tex]$) | 2.40 | 5.20 | 4.2 | The missing value for the total amount paid by Miss Ndlovu is \(4.2\). #### Define the variables Let: - \( x \) be the price per kilogram for Bananas. - \( y \) be the price per kilogram for Apples. #### Form separate equations using the information from each column Mrs Ndoro's Equation: \[ 1.2x + 0.8y = 2.40 \] Mrs Kori's Equation: \[ 2.2x + 2.0y = 5.20 \] Miss Ndlovu's Equation: \[ 1.8x + 1.6y = 4.2 \] ### PART B #### Solve for the unknown variables using the matrix method ##### Pulling out the matrix To solve the system of linear equations defined by Mrs Ndoro and Mrs Kori, we can write it in matrix form \( A \mathbf{x} = \mathbf{b} \). \[ A = \begin{pmatrix} 1.2 & 0.8 \\ 2.2 & 2.0 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} \] \[ \mathbf{b} = \begin{pmatrix} 2.40 \\ 5.20 \end{pmatrix} \] ##### Finding the determinant of the drawn matrix To find \( x \) and \( y \), we first need to check that the determinant of \( A \) is non-zero. \[ \text{Det}(A) = \begin{vmatrix} 1.2 & 0.8 \\ 2.2 & 2.0 \end{vmatrix} = (1.2 \cdot 2.0) - (0.8 \cdot 2.2) \] \[ \text{Det}(A) = 2.4 - 1.76 = 0.64 \] Since the determinant is \( 0.64 \), which is not zero, we can proceed to solve the system using the inverse of \( A \). #### Solving the system The inverse of matrix \( A \) is given by \[ A^{-1} = \frac{1}{\text{Det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] where \( a = 1.2 \), \( b = 0.8 \), \( c = 2.2 \), and \( d = 2.0 \). \[ A^{-1} = \frac{1}{0.64} \begin{pmatrix} 2.0 & -0.8 \\ -2.2 & 1.2 \end{pmatrix} = \begin{pmatrix} 3.125 & -1.25 \\ -3.4375 & 1.875 \end{pmatrix} \] Now we can find \( \mathbf{x} \) by multiplying \( A^{-1} \) with \( \mathbf{b} \): \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} 3.125 & -1.25 \\ -3.4375 & 1.875 \end{pmatrix} \begin{pmatrix} 2.40 \\ 5.20 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} (3.125 \cdot 2.40) + (-1.25 \cdot 5.20) \\ (-3.4375 \cdot 2.40) + (1.875 \cdot 5.20) \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} 7.5 - 6.5 \\ -8.25 + 9.75 \end{pmatrix} \] \[ \mathbf{x} = \begin{pmatrix} 1.00 \\ 1.50 \end{pmatrix} \] Thus, the price per kilogram for Bananas is \( \$[/tex]1.00 \) and the price per kilogram for Apples is [tex]\( \$1.50 \)[/tex].
Finally, we use these prices to calculate the total amount paid by Miss Ndlovu:
[tex]\[ 1.8x + 1.6y = 1.8 \cdot 1.00 + 1.6 \cdot 1.50 = 1.8 + 2.4 = 4.2 \][/tex]
So, Miss Ndlovu paid [tex]\( \$4.20 \)[/tex] for her fruits.
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