At Café A, a latte costs $3.50, a croissant costs $2.70 and a sandwich costs $5.00. At Café B, a latte costs $0.40 more, a croissant costs $0.20 less and a sandwich costs $0.50 less. This information can be represented by the matrix Q =

1. Arrange each buyer's purchases as rows in the order latte, croissant, sandwich.

Since $Q$ lists the items in the order latte, croissant, sandwich, the columns of $P$ should follow the same order.

Azza buys $2$ latte, $3$ croissants and $3$ sandwiches, so her row is
$(2\quad 3\quad 3).$

Yu Fei buys $1$ latte, $0$ croissants and $4$ sandwiches, so his row is
$(1\quad 0\quad 4).$

Therefore,
$P=\begin{pmatrix}2&3&3\\1&0&4\end{pmatrix}.$

1. Set up the product $PQ$ using the purchase matrix from part (a) and the given matrix $Q$.

$P=\begin{pmatrix}2&3&3\\1&0&4\end{pmatrix},\quad Q=\begin{pmatrix}3.50&0.40\\2.70&-0.20\\5.00&-0.50\end{pmatrix}$

$M=PQ=\begin{pmatrix}2&3&3\\1&0&4\end{pmatrix}\begin{pmatrix}3.50&0.40\\2.70&-0.20\\5.00&-0.50\end{pmatrix}$

2. Calculate each entry by multiplying the corresponding row of $P$ by each column of $Q$.

1. Identify Azza's totals from the first row of matrix $M$.

From part (b),
$M=\begin{pmatrix}28.10&-1.50\\23.50&-1.60\end{pmatrix}.$
Azza's row is the first row: $28.10$ and $-1.50$.

So Azza's total at Café A is $\char36{}28.10$, and the change to get Café B's total is $-\char36{}1.50$.

2. Find Azza's Café B total and decide which café is cheaper.

Azza's total at Café B is
$\$28.10-\$1.50=\$26.60.$
Since $\char36{}26.60<\char36{}28.10$, Café B is cheaper.