Answer :
Certainly! Let's go through each part step-by-step and solve it thoroughly.
### PART A
#### (a) Copy and complete the table
Given data:
- Mrs. Ndoro bought 2 kg of Pananas and 0.8 kg of Apples and paid [tex]$40. - Mrs. Kori bought 0.2 kg of Pananas and 0 kg of Apples and paid $[/tex]20.
- Miss Ndlovu bought 0.8 kg of Pananas and 6 kg of Apples, and we are to determine the total she paid.
Here is the completed table:
[tex]\[ \begin{array}{|l|c|c|c|} \hline & \text{Mrs Ndoro} & \text{Mrs Kori} & \text{Miss Ndlovu} \\ \hline \text{Pananas (kg)} & 2 & 0.2 & 0.8 \\ \text{Apples (kg)} & 0.8 & 0 & 6 \\ \hline \text{Total Paid (\$)} & 40 & 20 & ? \\ \hline \end{array} \][/tex]
#### (b) Define the variables
Let's define our variables:
- [tex]\( P \)[/tex] = Price per kilogram of Pananas
- [tex]\( A \)[/tex] = Price per kilogram of Apples
#### (c) Form separate equations using the information from each column
From the information provided, we can form the following equations:
1. For Mrs Ndoro:
[tex]\[ 2P + 0.8A = 40 \][/tex]
2. For Mrs Kori:
[tex]\[ 0.2P + 0A = 20 \][/tex]
3. For Miss Ndlovu:
[tex]\[ 0.8P + 6A = \text{Total Paid} \][/tex]
### PART B
#### (a) Pulling out the matrix
To apply the matrix method, we start by representing our system of equations in matrix form. For this, we only consider equations 1 and 2 to solve for [tex]\( P \)[/tex] and [tex]\( A \)[/tex].
The matrix representation is:
[tex]\[ \begin{pmatrix} 2 & 0.8 \\ 0.2 & 0 \end{pmatrix} \begin{pmatrix} P \\ A \end{pmatrix} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
#### (b) Finding the determinant of the drawn matrix
Let's compute the determinant of the coefficient matrix:
[tex]\[ \text{Determinant} = \begin{vmatrix} 2 & 0.8 \\ 0.2 & 0 \end{vmatrix} = (2 \cdot 0) - (0.8 \cdot 0.2) = 0 - 0.16 = -0.16 \][/tex]
Since the determinant is non-zero ([tex]\(-0.16\)[/tex]), we can solve the equations using the matrix method.
#### Solving the system of equations
Using matrix inversion or another appropriate methods to solve:
[tex]\[ \begin{pmatrix} P \\ A \end{pmatrix} = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
Calculating the inverse of the coefficient matrix and solving the system gives us:
[tex]\[ P \approx 100 \][/tex]
[tex]\[ A \approx -200 \][/tex]
Now, to find how much Miss Ndlovu paid:
[tex]\[ \text{Total Paid by Miss Ndlovu} = 0.8P + 6A \][/tex]
Substitute the values of [tex]\( P \)[/tex] and [tex]\( A \)[/tex]:
[tex]\[ \text{Total Paid by Miss Ndlovu} = 0.8 \times 100 + 6 \times (-200) = 80 - 1200 = -1120 \][/tex]
Therefore, the price per kilogram of Pananas is [tex]$100, the price per kilogram of Apples is $[/tex]-200, and Miss Ndlovu paid [tex]\(\$ -1120\)[/tex].
To summarize:
- [tex]\( P = 100 \)[/tex]
- [tex]\( A = -200 \)[/tex]
- Total Paid by Miss Ndlovu = [tex]\( -1120 \)[/tex]
- Determinant of the matrix = [tex]\( -0.16 \)[/tex]
Note: The negative price of apples is not realistic and indicates an inconsistency or error in the initial problem setup. However, as per our calculations with the given data, these are the results.
### PART A
#### (a) Copy and complete the table
Given data:
- Mrs. Ndoro bought 2 kg of Pananas and 0.8 kg of Apples and paid [tex]$40. - Mrs. Kori bought 0.2 kg of Pananas and 0 kg of Apples and paid $[/tex]20.
- Miss Ndlovu bought 0.8 kg of Pananas and 6 kg of Apples, and we are to determine the total she paid.
Here is the completed table:
[tex]\[ \begin{array}{|l|c|c|c|} \hline & \text{Mrs Ndoro} & \text{Mrs Kori} & \text{Miss Ndlovu} \\ \hline \text{Pananas (kg)} & 2 & 0.2 & 0.8 \\ \text{Apples (kg)} & 0.8 & 0 & 6 \\ \hline \text{Total Paid (\$)} & 40 & 20 & ? \\ \hline \end{array} \][/tex]
#### (b) Define the variables
Let's define our variables:
- [tex]\( P \)[/tex] = Price per kilogram of Pananas
- [tex]\( A \)[/tex] = Price per kilogram of Apples
#### (c) Form separate equations using the information from each column
From the information provided, we can form the following equations:
1. For Mrs Ndoro:
[tex]\[ 2P + 0.8A = 40 \][/tex]
2. For Mrs Kori:
[tex]\[ 0.2P + 0A = 20 \][/tex]
3. For Miss Ndlovu:
[tex]\[ 0.8P + 6A = \text{Total Paid} \][/tex]
### PART B
#### (a) Pulling out the matrix
To apply the matrix method, we start by representing our system of equations in matrix form. For this, we only consider equations 1 and 2 to solve for [tex]\( P \)[/tex] and [tex]\( A \)[/tex].
The matrix representation is:
[tex]\[ \begin{pmatrix} 2 & 0.8 \\ 0.2 & 0 \end{pmatrix} \begin{pmatrix} P \\ A \end{pmatrix} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
#### (b) Finding the determinant of the drawn matrix
Let's compute the determinant of the coefficient matrix:
[tex]\[ \text{Determinant} = \begin{vmatrix} 2 & 0.8 \\ 0.2 & 0 \end{vmatrix} = (2 \cdot 0) - (0.8 \cdot 0.2) = 0 - 0.16 = -0.16 \][/tex]
Since the determinant is non-zero ([tex]\(-0.16\)[/tex]), we can solve the equations using the matrix method.
#### Solving the system of equations
Using matrix inversion or another appropriate methods to solve:
[tex]\[ \begin{pmatrix} P \\ A \end{pmatrix} = \begin{pmatrix} 2 & 0.8 \\ 0.2 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 40 \\ 20 \end{pmatrix} \][/tex]
Calculating the inverse of the coefficient matrix and solving the system gives us:
[tex]\[ P \approx 100 \][/tex]
[tex]\[ A \approx -200 \][/tex]
Now, to find how much Miss Ndlovu paid:
[tex]\[ \text{Total Paid by Miss Ndlovu} = 0.8P + 6A \][/tex]
Substitute the values of [tex]\( P \)[/tex] and [tex]\( A \)[/tex]:
[tex]\[ \text{Total Paid by Miss Ndlovu} = 0.8 \times 100 + 6 \times (-200) = 80 - 1200 = -1120 \][/tex]
Therefore, the price per kilogram of Pananas is [tex]$100, the price per kilogram of Apples is $[/tex]-200, and Miss Ndlovu paid [tex]\(\$ -1120\)[/tex].
To summarize:
- [tex]\( P = 100 \)[/tex]
- [tex]\( A = -200 \)[/tex]
- Total Paid by Miss Ndlovu = [tex]\( -1120 \)[/tex]
- Determinant of the matrix = [tex]\( -0.16 \)[/tex]
Note: The negative price of apples is not realistic and indicates an inconsistency or error in the initial problem setup. However, as per our calculations with the given data, these are the results.
