Math problem involving matrices to calculate total scores for contestants based on points and difficulty.

Let's solve this step-by-step based on the problem presented in the image.

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Problem Overview

We are given two tables:

1. Points Awarded by judges for each contestant in three categories: Wall Hanging, Clothing, and Rug.
2. Degree of Difficulty for each category for each contestant.

To find the total score for each contestant:
> Total Score = (Points Awarded) × (Degree of Difficulty)

We are to:
- (a) Create matrices to organize the information.
- (b) Find the total score for each contestant using matrix multiplication.

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Step 1: Organize Data into Matrices

#### Matrix A: Points Awarded
This is a 3×3 matrix where rows represent contestants and columns represent categories.

$$
A = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14 & 17 \\
16 & 19.5 & 18
\end{bmatrix}
$$

Rows: Madison, Devyn, Ali
Columns: Wall Hanging, Clothing, Rug

---

#### Matrix D: Degree of Difficulty
This is a 3×3 matrix where rows represent categories and columns represent contestants.

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

Rows: Wall Hanging, Clothing, Rug
Columns: Madison, Devyn, Ali

---

⚠️ Important Note: To compute the total score for each contestant, we need to multiply the points awarded by the difficulty factor for each category.

So, for each contestant, we multiply their points in each category by the difficulty of that category, then sum them up.

That means we want:
> Total Score = $ A \times D^T $ ??

Wait — let’s think carefully.

Actually, the correct way is:

- For each contestant, we have:
- Wall Hanging score × Wall Hanging difficulty
- Clothing score × Clothing difficulty
- Rug score × Rug difficulty
- Sum them.

But the difficulty values are per category, not per contestant. So:

- The difficulty for Wall Hanging is the same for all contestants? Wait — looking at the table:

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

Ah! So the degree of difficulty varies per contestant — meaning it's not just a global difficulty per category, but each contestant has a different difficulty level for each category.

So the difficulty matrix D is correctly defined as:

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

Where:
- Row 1: Wall Hanging difficulty for Madison, Devyn, Ali
- Row 2: Clothing difficulty
- Row 3: Rug difficulty

Now, to compute the total score for each contestant, we do:

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

Similarly for others.

But we can do this using matrix multiplication.

We have:
- Matrix $ A $: 3×3 (Contestants × Categories)
- Matrix $ D $: 3×3 (Categories × Contestants)

Wait — dimensions don't align for direct multiplication unless we transpose.

But actually, to compute total score for each contestant, we want:

For each contestant $ i $, total score =
$$
\sum_{j=1}^{3} (\text{Points}_{i,j}) \times (\text{Difficulty}_{j,i})
$$

Wait — no. Let's reindex.

Let’s define:

Let $ P $ be the points matrix (3×3): Rows = contestants, Columns = categories

Let $ D $ be the difficulty matrix (3×3): Rows = categories, Columns = contestants

Then, to get total score for each contestant, we compute:

$$
\text{Total Score}_i = \sum_{k=1}^{3} P_{i,k} \cdot D_{k,i}
$$

That is, for contestant $ i $, sum over categories $ k $: (points of contestant $ i $ in category $ k $) × (difficulty of category $ k $ for contestant $ i $)

So this is like a dot product between row $ i $ of $ P $ and column $ i $ of $ D $

But since $ D $ has contestants in columns, we need to extract the appropriate entries.

Alternatively, we can transpose $ D $ so that columns become rows.

Let’s define:

Let $ D^T $ be the transpose of $ D $, which will be:

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

Wait — no:

Original $ D $:

$$
D = \begin{bmatrix}
2 & 3 & 2 \\ % Wall Hanging: M, D, A
3 & 3 & 1 \\ % Clothing
2 & 2 & 1 \\ % Rug
\end{bmatrix}
$$

So $ D^T $ is:

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

Now, $ D^T $ has:
- Rows: Madison, Devyn, Ali
- Columns: Wall Hanging, Clothing, Rug

So now both $ A $ and $ D^T $ are 3×3 with same row/column meanings.

But wait: $ A $ has:
- Rows: M, D, A
- Columns: Wall, Cloth, Rug

$ D^T $ also has:
- Rows: M, D, A
- Columns: Wall, Cloth, Rug

So now, element-wise multiplication is possible!

But we cannot just multiply matrices unless we use element-wise or dot product.

But we want:
For each contestant $ i $, total score = $ \sum_{j} A_{i,j} \cdot D^T_{i,j} $

That is, element-wise multiplication followed by row sum.

So we can compute:

$$
S = A \odot D^T
$$

Then sum each row.

But since the problem says "multiply the points awarded by the degree of difficulty", and the difficulty is per contestant per category, yes, we must multiply corresponding entries.

So here's the plan:

✔ Correct Approach:

Let’s define:
- $ A $: 3×3 matrix of points (contestant × category)
- $ D $: 3×3 matrix of difficulty (category × contestant)

We want:
- For each contestant $ i $, total score = $ \sum_{k=1}^{3} A_{i,k} \cdot D_{k,i} $

This is equivalent to computing the dot product of row $ i $ of $ A $ with column $ i $ of $ D $

But this is not standard matrix multiplication.

Instead, define $ D^T $ as above, then:

$$
\text{Score}_i = \sum_{j=1}^{3} A_{i,j} \cdot (D^T)_{i,j}
$$

So we can compute a new matrix $ S = A \odot D^T $, then sum each row.

But since this is likely expected to be done via matrix multiplication, perhaps there's a better way.

Wait — let's re-express.

Let’s write $ D $ as:

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

We want:
For Madison: $ 16.5 \cdot 2 + 18 \cdot 3 + 17.5 \cdot 2 $
For Devyn: $ 12.5 \cdot 3 + 14 \cdot 3 + 17 \cdot 2 $
For Ali: $ 16 \cdot 2 + 19.5 \cdot 1 + 18 \cdot 1 $

Wait — look at the difficulty values:

For Madison:
- Wall: 2
- Clothing: 3
- Rug: 2

So Madison’s difficulty vector is: [2, 3, 2]

Similarly:
- Devyn: [3, 3, 2]
- Ali: [2, 1, 1]

But from the table:
- Wall Hanging: Madison=2, Devyn=3, Ali=2 → so wall difficulty depends on contestant
- Clothing: Madison=3, Devyn=3, Ali=1
- Rug: Madison=2, Devyn=2, Ali=1

So yes, difficulty is per contestant per category.

So we can create a matrix $ D $ where:
- Rows: categories
- Columns: contestants

As already defined.

But to compute total score, we need:

$$
\text{Total Score} = A \cdot D^T
$$

Wait — let’s check dimensions:

- $ A $: 3×3 (contestants × categories)
- $ D^T $: 3×3 (contestants × categories)

No — $ D $ is 3×3 (categories × contestants), so $ D^T $ is 3×3 (contestants × categories)

But if we do $ A \cdot D^T $, we get:

- $ A $: 3×3
- $ D^T $: 3×3
- Result: 3×3

But what does this mean?

Let’s see:

Entry $ (i,j) $ of $ A \cdot D^T $ is:
$$
\sum_{k=1}^{3} A_{i,k} \cdot (D^T)_{k,j} = \sum_{k=1}^{3} A_{i,k} \cdot D_{j,k}
$$

That is: for contestant $ i $, category $ j $, sum over categories $ k $: points of $ i $ in $ k $ × difficulty of $ j $ in $ k $

That doesn’t make sense.

We want:
For contestant $ i $, total score = $ \sum_{k=1}^{3} A_{i,k} \cdot D_{k,i} $

That is: sum over categories $ k $: points of $ i $ in $ k $ × difficulty of $ k $ for $ i $

So it's $ \sum_{k} A_{i,k} \cdot D_{k,i} $

This is the trace of $ A \cdot D^T $? No.

It's the dot product of row $ i $ of $ A $ and column $ i $ of $ D $

So it's the diagonal elements of $ A \cdot D^T $? Let's see.

Let’s compute $ B = A \cdot D^T $

Then $ B_{i,i} = \sum_{k=1}^{3} A_{i,k} \cdot (D^T)_{k,i} = \sum_{k=1}^{3} A_{i,k} \cdot D_{i,k} $

Wait — no: $ (D^T)_{k,i} = D_{i,k} $

So $ B_{i,i} = \sum_{k=1}^{3} A_{i,k} \cdot D_{i,k} $

Yes! That’s exactly what we want.

But only if $ D $ is defined such that $ D_{i,k} $ is the difficulty of contestant $ i $ in category $ k $.

But in our current $ D $, we have:

- $ D_{1,1} = 2 $: Wall Hanging difficulty for Madison
- $ D_{1,2} = 3 $: Wall Hanging difficulty for Devyn
- etc.

So $ D_{i,j} $ = difficulty of category $ i $ for contestant $ j $

So $ D_{i,j} $ = difficulty of category $ i $ for contestant $ j $

Therefore, $ D^T_{j,i} = D_{i,j} $ = difficulty of category $ i $ for contestant $ j $

So $ (D^T)_{j,i} $ = difficulty of category $ i $ for contestant $ j $

But we want for contestant $ j $, sum over categories $ i $: $ A_{j,i} \cdot D_{i,j} $

Which is $ \sum_{i} A_{j,i} \cdot D_{i,j} $

Now, $ D_{i,j} = $ difficulty of category $ i $ for contestant $ j $

So yes.

Now, $ A_{j,i} $ = points of contestant $ j $ in category $ i $

So $ \sum_{i} A_{j,i} \cdot D_{i,j} $ = total score for contestant $ j $

But this is not a matrix product unless we reorganize.

Let’s define:

Let $ A $ be 3×3: $ A_{j,i} $ = points of contestant $ j $ in category $ i $

Let $ D $ be 3×3: $ D_{i,j} $ = difficulty of category $ i $ for contestant $ j $

Then the total score for contestant $ j $ is:
$$
\sum_{i=1}^{3} A_{j,i} \cdot D_{i,j}
$$

This is the dot product of row $ j $ of $ A $ and column $ j $ of $ D $

This is equal to the diagonal elements of $ A \cdot D^T $? Let's see.

Compute $ C = A \cdot D^T $

Then $ C_{j,k} = \sum_{i=1}^{3} A_{j,i} \cdot (D^T)_{i,k} = \sum_{i=1}^{3} A_{j,i} \cdot D_{k,i} $

So $ C_{j,j} = \sum_{i=1}^{3} A_{j,i} \cdot D_{j,i} $

But $ D_{j,i} $ = difficulty of category $ j $ for contestant $ i $

Not what we want.

We want $ \sum_{i} A_{j,i} \cdot D_{i,j} $

So it's $ \sum_{i} A_{j,i} \cdot D_{i,j} $

This is the dot product of row $ j $ of $ A $ and column $ j $ of $ D $

So it's the trace of $ A \cdot D^T $? No.

Let’s try defining a matrix $ E $ such that $ E_{i,j} = D_{i,j} $, then the desired sum is $ \sum_i A_{j,i} E_{i,j} $

This is the dot product of row $ j $ of $ A $ and column $ j $ of $ E $

So it's the diagonal of $ A \cdot E $ only if $ E $ is transposed.

Wait — let’s define $ F = D^T $, then $ F_{j,i} = D_{i,j} $

Then $ \sum_i A_{j,i} \cdot F_{j,i} = \sum_i A_{j,i} \cdot D_{i,j} $

So it’s the element-wise product of row $ j $ of $ A $ and row $ j $ of $ F $, then sum.

So we can compute $ S_j = \sum_{i} A_{j,i} \cdot F_{j,i} = \sum_{i} A_{j,i} \cdot D_{i,j} $

So we need to compute:

Let $ F = D^T $, then $ F = \begin{bmatrix} 2 & 3 & 2 \\ 3 & 3 & 2 \\ 2 & 1 & 1 \end{bmatrix} $

Now, $ A = \begin{bmatrix} 16.5 & 18 & 17.5 \\ 12.5 & 14 & 17 \\ 16 & 19.5 & 18 \end{bmatrix} $

Now, compute element-wise product $ A \odot F $, then sum each row.

Let’s do it manually.

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Part (b): Compute Total Scores

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

#### Devyn (row 2):
- Wall: 12.5 × 3 = 37.5
- Clothing: 14 × 3 = 42.0
- Rug: 17 × 2 = 34.0
- Total = 37.5 + 42 + 34 = 113.5

#### Ali (row 3):
- Wall: 16 × 2 = 32.0
- Clothing: 19.5 × 1 = 19.5
- Rug: 18 × 1 = 18.0
- Total = 32 + 19.5 + 18 = 69.5

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✔ Final Answers:

#### (a) Matrices:

- Points Awarded Matrix (A):
$$
A = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14 & 17 \\
16 & 19.5 & 18
\end{bmatrix}
$$

- Difficulty Matrix (D):
$$
D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$

(With rows: Wall, Clothing, Rug; columns: Madison, Devyn, Ali)

#### (b) Total Scores:

- Madison: 122.0
- Devyn: 113.5
- Ali: 69.5

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✔ Explanation:

The total score for each contestant is computed by multiplying the points awarded in each category by the degree of difficulty for that category for that contestant, then summing across all categories.

This is done element-wise multiplication of the points matrix and the transposed difficulty matrix, then summing each row.

Alternatively, it's the dot product of each contestant's point vector with their corresponding difficulty vector.

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🟩 Final Answer:

a. Matrices:

$$
\text{Points } A = \begin{bmatrix}
16.5 & 18 & 17.5 \\
12.5 & 14 & 17 \\
16 & 19.5 & 18
\end{bmatrix}, \quad
\text{Difficulty } D = \begin{bmatrix}
2 & 3 & 2 \\
3 & 3 & 1 \\
2 & 2 & 1
\end{bmatrix}
$$

b. Total Scores:

- Madison: $ 16.5 \times 2 + 18 \times 3 + 17.5 \times 2 = 33 + 54 + 35 = \boxed{122.0} $
- Devyn: $ 12.5 \times 3 + 14 \times 3 + 17 \times 2 = 37.5 + 42 + 34 = \boxed{113.5} $
- Ali: $ 16 \times 2 + 19.5 \times 1 + 18 \times 1 = 32 + 19.5 + 18 = \boxed{69.5} $

✔ These are the final scores.