Matrix Multiplication Word Problems Worksheet

Let's solve both problems step by step using matrix multiplication.

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Problem 1: Track and Field

We are given:
- Points for each place:
- 1st place: 5 points
- 2nd place: 3 points
- 3rd place: 1 point

We have a table showing how many times each school finished in each place:

| School | 1st Place | 2nd Place | 3rd Place |
|--------------|-----------|-----------|-----------|
| Jefferson | 8 | 4 | 5 |
| London | 6 | 3 | 7 |
| Springfield | 5 | 7 | 3 |
| Madison | 7 | 5 | 4 |

We need to compute the total points for each school.

#### Step 1: Represent the data as matrices

Let’s define:
- A points matrix (row vector):
$$
P = \begin{bmatrix} 5 & 3 & 1 \end{bmatrix}
$$

- A results matrix (each row is a school, columns are places):
$$
R = \begin{bmatrix}
8 & 4 & 5 \\
6 & 3 & 7 \\
5 & 7 & 3 \\
7 & 5 & 4
\end{bmatrix}
$$

Now, multiply $ P \times R $ to get total points per school.

$$
\text{Total Points} = P \cdot R = \begin{bmatrix} 5 & 3 & 1 \end{bmatrix} \cdot
\begin{bmatrix}
8 & 4 & 5 \\
6 & 3 & 7 \\
5 & 7 & 3
\end{bmatrix}
$$

Wait — actually, we should multiply the points vector with each row of the results matrix. So better to compute:

For each school:
$$
\text{Points} = (\text{1st place count}) \times 5 + (\text{2nd place count}) \times 3 + (\text{3rd place count}) \times 1
$$

So let's compute one by one:

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Jefferson:
$ 8 \times 5 + 4 \times 3 + 5 \times 1 = 40 + 12 + 5 = 57 $

London:
$ 6 \times 5 + 3 \times 3 + 7 \times 1 = 30 + 9 + 7 = 46 $

Springfield:
$ 5 \times 5 + 7 \times 3 + 3 \times 1 = 25 + 21 + 3 = 49 $

Madison:
$ 7 \times 5 + 5 \times 3 + 4 \times 1 = 35 + 15 + 4 = 54 $

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Now, let's write this as matrix multiplication:

Let’s define:
- $ P = \begin{bmatrix} 5 \\ 3 \\ 1 \end{bmatrix} $ (column vector of points)
- $ R^T $: transpose of results matrix so that each column is a school

But easier way: Multiply the row vector of points $ [5, 3, 1] $ with each row of the results matrix.

So:

$$
\text{Total Points} = \begin{bmatrix} 5 & 3 & 1 \end{bmatrix} \cdot \begin{bmatrix}
8 & 4 & 5 \\
6 & 3 & 7 \\
5 & 7 & 3 \\
7 & 5 & 4
\end{bmatrix}
= \begin{bmatrix}
5\cdot8 + 3\cdot4 + 1\cdot5 \\
5\cdot6 + 3\cdot3 + 1\cdot7 \\
5\cdot5 + 3\cdot7 + 1\cdot3 \\
5\cdot7 + 3\cdot5 + 1\cdot4
\end{bmatrix}
= \begin{bmatrix}
40+12+5 \\
30+9+7 \\
25+21+3 \\
35+15+4
\end{bmatrix}
= \begin{bmatrix}
57 \\
46 \\
49 \\
54
\end{bmatrix}
$$

So total points:
- Jefferson: 57
- London: 46
- Springfield: 49
- Madison: 54

✔ Jefferson won the meet with 57 points.

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✔ Answer to Problem 1:

- Jefferson: 57 points
- London: 46 points
- Springfield: 49 points
- Madison: 54 points
- Winner: Jefferson

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Problem 2: Chatsworth Baseball and Softball League

Given:
- Fees:
- Ages 7–8: $45
- Ages 9–10: $55
- Ages 11–14: $65

Number of players:

| Age | Baseball | Softball |
|---------|----------|----------|
| 7–8 | 350 | 280 |
| 9–10 | 320 | 165 |
| 11–14 | 180 | 120 |

We need to find the total money received.

#### Step 1: Write matrices

Let’s define:
- Fees matrix (column vector):
$$
F = \begin{bmatrix} 45 \\ 55 \\ 65 \end{bmatrix}
$$

- Players matrix (rows: age groups, columns: baseball and softball):
$$
P = \begin{bmatrix}
350 & 280 \\
320 & 165 \\
180 & 120
\end{bmatrix}
$$

We want to compute total revenue from all players.

Each player contributes their fee × number of players.

So, we can compute:

$$
\text{Total Revenue} = F^T \cdot P
$$

Where $ F^T = \begin{bmatrix} 45 & 55 & 65 \end{bmatrix} $

Then:
$$
\text{Revenue} = \begin{bmatrix} 45 & 55 & 65 \end{bmatrix} \cdot
\begin{bmatrix}
350 & 280 \\
320 & 165 \\
180 & 120
\end{bmatrix}
$$

This gives a 1×2 matrix (total revenue for baseball and softball).

Compute:

Baseball total:
$$
45 \cdot 350 + 55 \cdot 320 + 65 \cdot 180
$$

Calculate:
- $ 45 \times 350 = 15,750 $
- $ 55 \times 320 = 17,600 $
- $ 65 \times 180 = 11,700 $

Sum: $ 15,750 + 17,600 + 11,700 = 45,050 $

Softball total:
$$
45 \cdot 280 + 55 \cdot 165 + 65 \cdot 120
$$

Calculate:
- $ 45 \times 280 = 12,600 $
- $ 55 \times 165 = 9,075 $
- $ 65 \times 120 = 7,800 $

Sum: $ 12,600 + 9,075 + 7,800 = 29,475 $

Now, total league revenue:
$$
45,050 + 29,475 = \boxed{74,525}
$$

Alternatively, we could compute total revenue directly by multiplying fees with total players per age group.

Let’s verify:

- 7–8: $ (350 + 280) \times 45 = 630 \times 45 = 28,350 $
- 9–10: $ (320 + 165) \times 55 = 485 \times 55 = 26,675 $
- 11–14: $ (180 + 120) \times 65 = 300 \times 65 = 19,500 $

Total: $ 28,350 + 26,675 + 19,500 = 74,525 $

✔ Same result.

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✔ Answer to Problem 2:

- Total revenue: $74,525

Matrix representation:
- Fees matrix: $ \begin{bmatrix} 45 \\ 55 \\ 65 \end{bmatrix} $
- Players matrix: $ \begin{bmatrix} 350 & 280 \\ 320 & 165 \\ 180 & 120 \end{bmatrix} $
- Total revenue: $ \begin{bmatrix} 45 & 55 & 65 \end{bmatrix} \cdot \begin{bmatrix} 350 & 280 \\ 320 & 165 \\ 180 & 120 \end{bmatrix} = \begin{bmatrix} 45,050 & 29,475 \end{bmatrix} $
- Total money: $ 45,050 + 29,475 = \boxed{\$74,525} $

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✔ Final Answers:

1. Track and Field:
- Jefferson: 57 points
- London: 46 points
- Springfield: 49 points
- Madison: 54 points
- Winner: Jefferson

2. Chatsworth League:
- Total revenue: \$74,525

Let me know if you'd like these written in a formatted answer sheet!