Worksheet with matrix multiplication word problems on flower sales, competition scores, and store revenue.

Let's solve each problem step by step using matrix multiplication and linear systems, as requested. We'll show all work and explain the reasoning.

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Problem 1: Flower Sale

#### Given:
- Bought:
- 200 roses at $1.67 each
- 150 daffodils at $1.03 each
- 100 orchids at $2.59 each

- Sold:
- Roses for $3.00 each
- Daffodils for $2.25 each
- Orchids for $4.50 each

---

#### a) Total amount spent on flowers (cost)

We need to organize data into two matrices and use matrix multiplication.

Let’s define:

- Matrix A (Quantity bought):
This is a row vector representing quantities of each flower.
$$
A = \begin{bmatrix} 200 & 150 & 100 \end{bmatrix}
$$

- Matrix B (Cost per unit):
This is a column vector representing cost per flower.
$$
B = \begin{bmatrix} 1.67 \\ 1.03 \\ 2.59 \end{bmatrix}
$$

Now multiply:
$$
\text{Total Cost} = A \times B = \begin{bmatrix} 200 & 150 & 100 \end{bmatrix} \begin{bmatrix} 1.67 \\ 1.03 \\ 2.59 \end{bmatrix}
$$

Compute:
$$
= 200(1.67) + 150(1.03) + 100(2.59)
= 334 + 154.5 + 259 = 747.50
$$

✔ Total amount spent: $747.50

---

#### b) Total amount received from sale (revenue)

Now we use selling prices.

- Matrix C (Selling price per unit):
$$
C = \begin{bmatrix} 3.00 \\ 2.25 \\ 4.50 \end{bmatrix}
$$

Use same quantity matrix $ A = \begin{bmatrix} 200 & 150 & 100 \end{bmatrix} $

$$
\text{Revenue} = A \times C = 200(3.00) + 150(2.25) + 100(4.50)
= 600 + 337.5 + 450 = 1387.50
$$

✔ Total revenue received: $1,387.50

---

#### c) Profit made

Use matrix operations:
$$
\text{Profit} = \text{Revenue} - \text{Cost} = 1387.50 - 747.50 = 640.00
$$

Alternatively, we can create a profit per unit matrix and multiply:

- Profit per unit: Selling price – Cost price
$$
\begin{bmatrix} 3.00 - 1.67 \\ 2.25 - 1.03 \\ 4.50 - 2.59 \end{bmatrix}
= \begin{bmatrix} 1.33 \\ 1.22 \\ 1.91 \end{bmatrix}
$$

Then:
$$
\text{Profit} = \begin{bmatrix} 200 & 150 & 100 \end{bmatrix} \begin{bmatrix} 1.33 \\ 1.22 \\ 1.91 \end{bmatrix}
= 200(1.33) + 150(1.22) + 100(1.91)
= 266 + 183 + 191 = 640
$$

✔ Total profit: $640.00

---

✔ Answer Summary for Problem 1:

- a) Total spent: $747.50
- b) Total received: $1,387.50
- c) Profit: $640.00

---

Problem 2: Sheep and Wool Fair Scores

#### Given:
- Points awarded by judges for each contestant in 3 categories.
- Degree of difficulty for each category per contestant.

We are to compute total score for each contestant.

> Note: The total score is likely points × difficulty, so we need to weight the points by difficulty.

But wait — let's interpret carefully.

The table shows:
- Each contestant got points for each category.
- Each category has a difficulty multiplier for each contestant.

So the total score for a contestant in a category = Points × Difficulty factor

Thus, we can model this with matrix multiplication.

Let’s define:

#### a) Create matrices

Matrix P (Points awarded):
Rows = Contestants, Columns = Categories

$$
P = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14.0 & 17.0 \\
16.0 & 19.5 & 18.0
\end{bmatrix}
$$

Matrix D (Difficulty factors):
Rows = Categories, Columns = Contestants

$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$

Wait — now we have:
- $ P $: 3×3 (contestants × categories)
- $ D $: 3×3 (categories × contestants)

To compute total score for each contestant, we need:
$$
\text{Score}_i = \sum_{j} (\text{Points}_{ij} \times \text{Difficulty}_{ji})
$$

This suggests we should transpose one matrix.

But actually, we want:
For Madison:
- Wall Hanging: 16.5 × 2 = 33.0
- Clothing: 18 × 3 = 54.0
- Rug: 17.5 × 2 = 35.0
→ Total = 33 + 54 + 35 = 122.0

So we need to multiply each row of Points by the corresponding column of Difficulty.

So we can do:
$$
\text{Total Score} = P \times D^T
$$

Wait — better way: Let’s reorganize.

Let’s define:

Matrix D (transposed): Now columns are contestants, rows are categories → matches P

Actually, if we keep D as is:
- D is Category × Contestant, so it’s 3×3

We want to compute:
$$
\text{Score} = P \times D
$$

But dimensions:
- $ P $: 3×3
- $ D $: 3×3 → valid

But check:
- Row i of P (contestant i’s points) × Column j of D (difficulty for contestant j)? That doesn’t make sense.

Ah! There's a mismatch.

Wait — look again:

Difficulty table:

| Category | Madison | Devyn | Ali |
|--------------|---------|-------|-----|
| Wall Hanging | 2 | 3 | 2 |
| Clothing | 3 | 3 | 1 |
| Rug | 2 | 2 | 1 |

So each contestant has a difficulty value for each category.

So for Madison, her difficulty values are:
- Wall Hanging: 2
- Clothing: 3
- Rug: 2

So we can write Difficulty matrix D as:
$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$
with rows = categories, columns = contestants

And Points matrix P:
$$
P = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14.0 & 17.0 \\
16.0 & 19.5 & 18.0
\end{bmatrix}
$$
rows = contestants, columns = categories

So to get total score for each contestant, we need:
$$
\text{Score}_i = \sum_{\text{category}} (\text{Points}_{i,\text{cat}} \times \text{Difficulty}_{\text{cat},i})
$$

So we can compute:
$$
\text{Score Matrix} = P \times D^T
$$

Wait — no.

Actually, if we want to multiply Points (contestant × category) with Difficulty (category × contestant), then:

We need to transpose D so that it becomes contestant × category, but that’s not right.

Better approach:

Let’s define D as a matrix where rows = contestants, columns = categories, same as P.

So transpose D:

$$
D^T = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}^T = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 2 \\
2 & 1 & 1
\end{bmatrix}
$$

Wait — no:

Original D:
$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
\Rightarrow \text{row 1: Madison's difficulties? No!}
$$

Wait — confusion!

Actually, the first column of D is Madison's difficulty for each category.

So:
- D column 1: Madison → [2, 3, 2]^T
- D column 2: Devyn → [3, 3, 2]^T
- D column 3: Ali → [2, 1, 1]^T

So D is 3×3, with columns = contestants, rows = categories

So D is:
$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$

Now, P is 3×3, with rows = contestants, columns = categories

So to compute total score for each contestant, we need:
$$
\text{Score}_i = \sum_{j=1}^{3} P_{i,j} \cdot D_{j,i}
$$

That is, element-wise product across categories, then sum.

This is not standard matrix multiplication unless we transpose.

So define:
- $ P $: 3×3 (contestant × category)
- $ D^T $: 3×3 (category × contestant) → now rows = categories, columns = contestants

Then:
$$
\text{Score} = P \times D^T
$$

Yes! Because:
- $ P $: contestant × category
- $ D^T $: category × contestant
- Result: contestant × contestant → no!

Wait — we want a vector of scores per contestant.

So better: define D as a column vector per contestant.

Let’s do it one contestant at a time, or use matrix multiplication properly.

Let’s define:

Let D be a 3×3 matrix where:
- Rows = categories
- Columns = contestants

So:
$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$

Now, P is:
$$
P = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14.0 & 17.0 \\
16.0 & 19.5 & 18.0
\end{bmatrix}
$$
Rows = contestants, columns = categories

Now, we want:
For contestant i:
$$
\text{Score}_i = \sum_{k=1}^3 P_{i,k} \cdot D_{k,i}
$$

This is the dot product of row i of P with column i of D.

So we can compute:
$$
\text{Score} = \text{diag}(P \cdot D^T)
$$

But easier: just compute manually.

Let’s do it:

#### For Madison (i=1):
- Wall: 16.5 × 2 = 33.0
- Clothing: 18 × 3 = 54.0
- Rug: 17.5 × 2 = 35.0
- Total = 33 + 54 + 35 = 122.0

#### For Devyn (i=2):
- Wall: 12.5 × 3 = 37.5
- Clothing: 14.0 × 3 = 42.0
- Rug: 17.0 × 2 = 34.0
- Total = 37.5 + 42 + 34 = 113.5

#### For Ali (i=3):
- Wall: 16.0 × 2 = 32.0
- Clothing: 19.5 × 1 = 19.5
- Rug: 18.0 × 1 = 18.0
- Total = 32 + 19.5 + 18 = 69.5

✔ So total scores:
- Madison: 122.0
- Devyn: 113.5
- Ali: 69.5

---

✔ Answer for Problem 2:

- a) Matrices:
$$
P = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14.0 & 17.0 \\
16.0 & 19.5 & 18.0
\end{bmatrix}, \quad
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$
- b) Total scores:
- Madison: 122.0
- Devyn: 113.5
- Ali: 69.5

---

Problem 3: Hardware Store Revenue

#### Given:
- Prices:
- Hammer: $3.00
- Flashlight: $5.00
- Lantern: $7.00

- Sales:
- Store A: 10 hammers, 2 flashlights, 2 lanterns
- Store B: 9 hammers, 14 flashlights, 5 lanterns
- Store C: 8 hammers, 6 flashlights, 7 lanterns

---

#### a) Create matrices

Price matrix (P): 1×3 row matrix
$$
P = \begin{bmatrix} 3.00 & 5.00 & 7.00 \end{bmatrix}
$$

Sales matrix (S): 3×3 matrix (stores × items)
$$
S = \begin{bmatrix}
10 & 2 & 2 \\
9 & 14 & 5 \\
8 & 6 & 7
\end{bmatrix}
$$
Rows = stores, columns = items

---

#### b) Product of matrices

Compute:
$$
R = S \times P^T
$$

Wait — dimensions:
- $ S $: 3×3
- $ P $: 1×3 → so $ P^T $: 3×1

So:
$$
R = S \times P^T = \begin{bmatrix}
10 & 2 & 2 \\
9 & 14 & 5 \\
8 & 6 & 7
\end{bmatrix}
\times
\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}
$$

Result: 3×1 matrix → revenue per store.

Compute:

- Store A: $10(3) + 2(5) + 2(7) = 30 + 10 + 14 = 54$
- Store B: $9(3) + 14(5) + 5(7) = 27 + 70 + 35 = 132$
- Store C: $8(3) + 6(5) + 7(7) = 24 + 30 + 49 = 103$

So:
$$
R = \begin{bmatrix} 54 \\ 132 \\ 103 \end{bmatrix}
$$

✔ The product represents the gross revenue for each store.

---

#### c) How to find total gross revenue from all three stores?

Add up the revenues:
$$
54 + 132 + 103 = 289
$$

Or, use matrix sum:
$$
\text{Total Revenue} = \sum_{i=1}^3 R_i = 289
$$

---

#### d) Total gross revenue from flashlights sold at all stores

Flashlight price: $5.00

Flashlights sold:
- Store A: 2
- Store B: 14
- Store C: 6
→ Total = 2 + 14 + 6 = 22

Revenue: $22 × 5 = \$110$

Using matrix: extract flashlight column from S: $[2, 14, 6]^T$, multiply by $5:
$$
\begin{bmatrix} 2 \\ 14 \\ 6 \end{bmatrix} \times 5 = 110
$$

✔ Total flashlight revenue: $110

---

✔ Answer Summary for Problem 3:

- a)
$$
P = \begin{bmatrix} 3 & 5 & 7 \end{bmatrix}, \quad
S = \begin{bmatrix}
10 & 2 & 2 \\
9 & 14 & 5 \\
8 & 6 & 7
\end{bmatrix}
$$
- b) Product $ S \times P^T $ gives revenue per store.
- c) Total revenue: $54 + 132 + 103 = \$289$
- d) Flashlight revenue: $110

---

✔ Final Answers:

#### Problem 1:
- a) Total spent: $747.50
- b) Total received: $1,387.50
- c) Profit: $640.00

#### Problem 2:
- a) Matrices as above
- b) Scores:
- Madison: 122.0
- Devyn: 113.5
- Ali: 69.5

#### Problem 3:
- a) Price and sales matrices defined
- b) Product gives revenue per store
- c) Total revenue: $289
- d) Flashlight revenue: $110

All problems solved using matrix multiplication and clear reasoning.