Answer :
Let's break down the entire problem step by step, translating the information into a structured mathematical solution.
## PART A
### Given Data Table
Given the data table, it looks something like this:
[tex]\[ \begin{array}{|l|c|c|c|} \hline & \text{Mrs Ndoro} & \text{Mrs Kori} & \text{Miss Ndlovu} \\ \hline \text{Bananas (kg)} & 2 & 0.2 & 0.8 \\ \text{Apples (kg)} & 0.8 & 1 & 6 \\ \hline \text{Total Paid (\$)} & 40 & 20 & T \\ \hline \end{array} \][/tex]
### (a) Copy and Complete the Table
We are given that [tex]\( T \)[/tex] is the total amount paid by Miss Ndlovu, which we will calculate later.
### (b) Define the Variables
Let's define our variables:
- Let [tex]\( x \)[/tex] be the price per kilogram (kg) of bananas.
- Let [tex]\( y \)[/tex] be the price per kilogram (kg) of apples.
### (c) Form Separate Equations
Given the weights and total amounts paid by Mrs Ndoro, Mrs Kori, and Miss Ndlovu, we can form the following equations:
1. For Mrs Ndoro:
[tex]\[ 2x + 0.8y = 40 \][/tex]
2. For Mrs Kori:
[tex]\[ 0.2x + y = 20 \][/tex]
3. For Miss Ndlovu:
[tex]\[ 0.8x + 6y = T \][/tex]
## PART B
### (a) Pulling Out the Matrix
The system of equations can be represented in matrix form. The coefficient matrix (let's call it [tex]\( A \)[/tex]) and the constant matrix (let's call it [tex]\( B \)[/tex]) are extracted from the first two equations.
Matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 1 \\ 0.8 & 6 \end{pmatrix} \][/tex]
Matrix [tex]\( B \)[/tex]:
[tex]\[ B = \begin{pmatrix} 40 \\ 20 \\ T \end{pmatrix} \][/tex]
### Solving the System and Finding Determinant
To find the price per kilogram of bananas and apples ([tex]\( x \)[/tex] and [tex]\( y \)[/tex]), we take the coefficient matrix from the first two equations and solve using the matrix method:
Sub-matrix [tex]\( A_{subst} \)[/tex]:
[tex]\[ A_{subst} = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 1 \end{pmatrix} \][/tex]
Sub-matrix [tex]\( B_{subst} \)[/tex]:
[tex]\[ B_{subst} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] using matrix methods:
[tex]\[ \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}_{subst} B_{subst} \][/tex]
The solutions are:
- Price per kg of bananas ([tex]\( x \)[/tex]) = 13.043478260869563
- Price per kg of apples ([tex]\( y \)[/tex]) = 17.39130434782609
So, [tex]\( x \approx 13.04 \)[/tex] and [tex]\( y \approx 17.39 \)[/tex].
### Finding Total Paid by Miss Ndlovu
Using the obtained values [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the third equation:
[tex]\[ 0.8 x + 6 y = T \][/tex]
Substituting:
[tex]\[ 0.8 \times 13.04 + 6 \times 17.39 = T \][/tex]
This yields the total amount paid by Miss Ndlovu:
[tex]\[ T \approx 114.78 \][/tex]
### (b) Finding the Determinant of the Drawn Matrix
To find the determinant of the sub-matrix [tex]\( A_{subst} \)[/tex]:
[tex]\[ \text{det}\left( A_{subst} \right) = \left| \begin{matrix} 2 & 0.8 \\ 0.2 & 1 \end{matrix} \right| \][/tex]
[tex]\[ \text{det}\left( A_{subst} \right) = (2 \times 1) - (0.8 \times 0.2) \][/tex]
[tex]\[ \text{det}\left( A_{subst} \right) = 2 - 0.16 = 1.84 \][/tex]
So, the determinant of the sub-matrix is [tex]\( 1.84 \)[/tex].
## Summary
- Price per kg of bananas: [tex]\( 13.04 \)[/tex]
- Price per kg of apples: [tex]\( 17.39 \)[/tex]
- Total amount paid by Miss Ndlovu: [tex]\( 114.78 \)[/tex]
- Determinant of sub-matrix: [tex]\( 1.84 \)[/tex]
This solves the problem using matrices as required.
## PART A
### Given Data Table
Given the data table, it looks something like this:
[tex]\[ \begin{array}{|l|c|c|c|} \hline & \text{Mrs Ndoro} & \text{Mrs Kori} & \text{Miss Ndlovu} \\ \hline \text{Bananas (kg)} & 2 & 0.2 & 0.8 \\ \text{Apples (kg)} & 0.8 & 1 & 6 \\ \hline \text{Total Paid (\$)} & 40 & 20 & T \\ \hline \end{array} \][/tex]
### (a) Copy and Complete the Table
We are given that [tex]\( T \)[/tex] is the total amount paid by Miss Ndlovu, which we will calculate later.
### (b) Define the Variables
Let's define our variables:
- Let [tex]\( x \)[/tex] be the price per kilogram (kg) of bananas.
- Let [tex]\( y \)[/tex] be the price per kilogram (kg) of apples.
### (c) Form Separate Equations
Given the weights and total amounts paid by Mrs Ndoro, Mrs Kori, and Miss Ndlovu, we can form the following equations:
1. For Mrs Ndoro:
[tex]\[ 2x + 0.8y = 40 \][/tex]
2. For Mrs Kori:
[tex]\[ 0.2x + y = 20 \][/tex]
3. For Miss Ndlovu:
[tex]\[ 0.8x + 6y = T \][/tex]
## PART B
### (a) Pulling Out the Matrix
The system of equations can be represented in matrix form. The coefficient matrix (let's call it [tex]\( A \)[/tex]) and the constant matrix (let's call it [tex]\( B \)[/tex]) are extracted from the first two equations.
Matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 1 \\ 0.8 & 6 \end{pmatrix} \][/tex]
Matrix [tex]\( B \)[/tex]:
[tex]\[ B = \begin{pmatrix} 40 \\ 20 \\ T \end{pmatrix} \][/tex]
### Solving the System and Finding Determinant
To find the price per kilogram of bananas and apples ([tex]\( x \)[/tex] and [tex]\( y \)[/tex]), we take the coefficient matrix from the first two equations and solve using the matrix method:
Sub-matrix [tex]\( A_{subst} \)[/tex]:
[tex]\[ A_{subst} = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 1 \end{pmatrix} \][/tex]
Sub-matrix [tex]\( B_{subst} \)[/tex]:
[tex]\[ B_{subst} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] using matrix methods:
[tex]\[ \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}_{subst} B_{subst} \][/tex]
The solutions are:
- Price per kg of bananas ([tex]\( x \)[/tex]) = 13.043478260869563
- Price per kg of apples ([tex]\( y \)[/tex]) = 17.39130434782609
So, [tex]\( x \approx 13.04 \)[/tex] and [tex]\( y \approx 17.39 \)[/tex].
### Finding Total Paid by Miss Ndlovu
Using the obtained values [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the third equation:
[tex]\[ 0.8 x + 6 y = T \][/tex]
Substituting:
[tex]\[ 0.8 \times 13.04 + 6 \times 17.39 = T \][/tex]
This yields the total amount paid by Miss Ndlovu:
[tex]\[ T \approx 114.78 \][/tex]
### (b) Finding the Determinant of the Drawn Matrix
To find the determinant of the sub-matrix [tex]\( A_{subst} \)[/tex]:
[tex]\[ \text{det}\left( A_{subst} \right) = \left| \begin{matrix} 2 & 0.8 \\ 0.2 & 1 \end{matrix} \right| \][/tex]
[tex]\[ \text{det}\left( A_{subst} \right) = (2 \times 1) - (0.8 \times 0.2) \][/tex]
[tex]\[ \text{det}\left( A_{subst} \right) = 2 - 0.16 = 1.84 \][/tex]
So, the determinant of the sub-matrix is [tex]\( 1.84 \)[/tex].
## Summary
- Price per kg of bananas: [tex]\( 13.04 \)[/tex]
- Price per kg of apples: [tex]\( 17.39 \)[/tex]
- Total amount paid by Miss Ndlovu: [tex]\( 114.78 \)[/tex]
- Determinant of sub-matrix: [tex]\( 1.84 \)[/tex]
This solves the problem using matrices as required.
