The problem asks us to fill in the blanks to form a matrix equation that can be used to calculate the individual prices for each size of coffee. We are given the matrix $R$, which represents the takings from the sale of coffees for the first three days of the first week. We need to find a matrix equation that relates the number of coffees sold and their prices to the total takings. We also need to solve the equation from part i), stating the cost of each size of coffee.

1. Problem Description

The problem asks us to fill in the blanks to form a matrix equation that can be used to calculate the individual prices for each size of coffee. We are given the matrix $R$, which represents the takings from the sale of coffees for the first three days of the first week. We need to find a matrix equation that relates the number of coffees sold and their prices to the total takings. We also need to solve the equation from part i), stating the cost of each size of coffee.

2. Solution Steps

First, we need to understand the given information and what the question is asking. $R$ is a column matrix with entries $307, 224.6, 235.8$. These values correspond to the takings for Monday (M), Tuesday (T), and Wednesday (W), respectively.

The prices for the small (S), medium (M), and large (L) coffees are what we are trying to find. Let's call these prices $p_S, p_M,$ and $p_L$ respectively. Thus, we can represent the prices as a column matrix:

$\begin{bmatrix}

p_S \\

p_M \\

p_L

\end{bmatrix}$

The matrix we need to find represents the number of each size of coffee sold on each day. Let $a_{ij}$ be the number of coffees of size $j$ sold on day $i$, where $i$ represents the days (M, T, W) and $j$ represents the size of the coffees (S, M, L). Then the matrix representing the number of coffees sold is:

$\begin{bmatrix}

a_{MS} & a_{MM} & a_{ML} \\

a_{TS} & a_{TM} & a_{TL} \\

a_{WS} & a_{WM} & a_{WL}

\end{bmatrix}$

From the blanks, we know that the number of each coffee sold on each day are $a_{MS}=10, a_{MM}=20, a_{ML}=30, a_{TS}=12, a_{TM}=15, a_{TL}=20, a_{WS}=8, a_{WM}=12, a_{WL}=25$. In this case, the matrix given in the problem is

$\begin{bmatrix}

10 & 20 & 30 \\

12 & 15 & 20 \\

8 & 12 & 25

\end{bmatrix}$

The matrix equation can be written as

$\begin{bmatrix}

10 & 20 & 30 \\

12 & 15 & 20 \\

8 & 12 & 25

\end{bmatrix}

\begin{bmatrix}

p_S \\

p_M \\

p_L

\end{bmatrix}

=

\begin{bmatrix}

307 \\

224.6 \\

235.8

\end{bmatrix}$

Solving the equation above gives

$10p_S + 20p_M + 30p_L = 307$

$12p_S + 15p_M + 20p_L = 224.6$

$8p_S + 12p_M + 25p_L = 235.8$

Solving these system of equations:

Multiply the first equation by 6 and the second equation by 5:

$60p_S + 120p_M + 180p_L = 1842$

$60p_S + 75p_M + 100p_L = 1123$

Subtract the equations:

$45p_M + 80p_L = 719$

Multiply the first equation by 4 and the third equation by 5:

$40p_S + 80p_M + 120p_L = 1228$

$40p_S + 60p_M + 125p_L = 1179$

Subtract the equations:

$20p_M - 5p_L = 49$

Now we have

$45p_M + 80p_L = 719$

$20p_M - 5p_L = 49$

Multiply the second equation by 16:

$320p_M - 80p_L = 784$

Adding this equation to the first:

$365p_M = 1503$

$p_M = 4.117808219178082 \approx 4.12$

Plug into $20p_M - 5p_L = 49$

$20(4.12) - 5p_L = 49$

$82.4 - 5p_L = 49$

$5p_L = 33.4$

$p_L = 6.68$

Plug into $10p_S + 20p_M + 30p_L = 307$

$10p_S + 20(4.12) + 30(6.68) = 307$

$10p_S + 82.4 + 200.4 = 307$

$10p_S = 24.2$

$p_S = 2.42$

3. Final Answer

The matrix equation is:

$\begin{bmatrix}

10 & 20 & 30 \\

12 & 15 & 20 \\

8 & 12 & 25

\end{bmatrix}

\begin{bmatrix}

S \\

M \\

L

\end{bmatrix}

=

\begin{bmatrix}

307 \\

224.6 \\

235.8

\end{bmatrix}$

The prices are:

Small coffee: $2.42

Medium coffee: $4.12

Large coffee: $6.68