The problem presents the prices per kilogram for apples, pears, and bananas. Sean wants to buy 2 kg of apples, 1 kg of pears, and 3 kg of bananas. The task is to find which of the given matrix products will result in a $1 \times 1$ matrix that represents the total cost of Sean's purchase.

1. Problem Description

The problem presents the prices per kilogram for apples, pears, and bananas. Sean wants to buy 2 kg of apples, 1 kg of pears, and 3 kg of bananas. The task is to find which of the given matrix products will result in a $1 \times 1$ matrix that represents the total cost of Sean's purchase.

2. Solution Steps

First, let's calculate the total cost manually:

Total cost = (2 kg of apples * $2.50/kg) + (1 kg of pears *$3.20/kg) + (3 kg of bananas * $1.90/kg)

Total cost = $(2 \times 2.50) + (1 \times 3.20) + (3 \times 1.90) = 5.00 + 3.20 + 5.70 = 13.90$

Now, let's examine each option:

A.

$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix} = \begin{bmatrix} 2(2.50) + 0(3.20) + 0(1.90) \\ 0(2.50) + 1(3.20) + 0(1.90) \\ 0(2.50) + 0(3.20) + 3(1.90) \end{bmatrix} = \begin{bmatrix} 5.00 \\ 3.20 \\ 5.70 \end{bmatrix}$.

This is a $3 \times 1$ matrix, so this is incorrect.

B.

$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix} = 2(2.50) + 1(3.20) + 3(1.90) = 5.00 + 3.20 + 5.70 = 13.90$

This results in a $1 \times 1$ matrix.

C.

$\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix}$. The dimensions of the matrices are $3 \times 1$ and $3 \times 1$. For matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix. In this case it doesn't work because the matrices must be in the form $n \times m$ multiplied by $m \times k$. In our situation, we have $3 \times 1$ and $3 \times 1$, therefore it won't work. This is undefined.

D.

$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix} = 2(2.50) + 1(3.20) + 3(1.90) = 5.00 + 3.20 + 5.70 = 13.90$

This results in a $1 \times 1$ matrix.

Options B and D are equivalent and correct.

Part a) We will evaluate option B.

$\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix} = [2(2.50) + 1(3.20) + 3(1.90)] = [5.00 + 3.20 + 5.70] = [13.90]$

Part b)

From the explanation above, option C gives an undefined product matrix since the matrix multiplication cannot be performed.

3. Final Answer

a) B. $\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2.50 \\ 3.20 \\ 1.90 \end{bmatrix} = [13.90]$

b) C.