The problem provides matrices $A$ and $B$, where:

$A = \begin{pmatrix} 4 & 4 \\ 2 & 2 \\ 3 & 7 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & y \\ x & 2 \\ 3 & 7 \end{pmatrix}$

It states that $A^T = B^T$, where $A^T$ and $B^T$ are the transposes of $A$ and $B$, respectively. You are tasked with finding the value of $y^x$.

Step-by-Step Solution:

1. Transpose of $A$: $A^T = \begin{pmatrix} 4 & 2 & 3 \\ 4 & 2 & 7 \end{pmatrix}$

2. Transpose of $B$: $B^T = \begin{pmatrix} 4 & x & 3 \\ y & 2 & 7 \end{pmatrix}$

3. Equating $A^T$ and $B^T$: Since $A^T = B^T$, we compare corresponding elements of the matrices:

   • From the first row: $4 = 4$, $x = 2$, and $3 = 3$.
   • From the second row: $y = 4$, $2 = 2$, and $7 = 7$.

   Therefore: $x = 2, \quad y = 4$

4. Finding $y^x$: Substitute $y = 4$ and $x = 2$: $y^x = 4^2 = 16$

Final Answer:

$\boxed{16 \, (B)}$

Do you want further clarifications or additional examples?

Related Questions:

1. How do you compute the transpose of a matrix?
2. What are the properties of matrix transposition?
3. How can you verify if two matrices are equal?
4. How would this problem change if $A \neq B$?
5. What are practical applications of matrix transposition in linear algebra?

Tip:

Always compare corresponding elements carefully when working with matrix equality!