The problem requires us to find the product of two matrices $D$ and $E$, where $D = \begin{bmatrix} 2 & 6 \\ 7 & 3 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 7 & 1 & 0 \end{bmatrix}$.

1. Problem Description

The problem requires us to find the product of two matrices $D$ and $E$, where

$D = \begin{bmatrix} 2 & 6 \\ 7 & 3 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 7 & 1 & 0 \end{bmatrix}$.

2. Solution Steps

To find the product $DE$, we need to check if the number of columns in $D$ is equal to the number of rows in $E$.

$D$ is a $2 \times 2$ matrix, and $E$ is a $2 \times 3$ matrix.

Since the number of columns in $D$ (2) is equal to the number of rows in $E$ (2), the product $DE$ is defined and will be a $2 \times 3$ matrix.

To calculate the elements of $DE$, we use the following formula:

If $DE = C$, then $c_{ij} = \sum_{k=1}^{n} d_{ik}e_{kj}$, where $n$ is the number of columns in $D$ (and the number of rows in $E$).

$c_{11} = (2)(1) + (6)(7) = 2 + 42 = 44$

$c_{12} = (2)(0) + (6)(1) = 0 + 6 = 6$

$c_{13} = (2)(3) + (6)(0) = 6 + 0 = 6$

$c_{21} = (7)(1) + (3)(7) = 7 + 21 = 28$

$c_{22} = (7)(0) + (3)(1) = 0 + 3 = 3$

$c_{23} = (7)(3) + (3)(0) = 21 + 0 = 21$

Therefore, $DE = \begin{bmatrix} 44 & 6 & 6 \\ 28 & 3 & 21 \end{bmatrix}$.

3. Final Answer

$DE = \begin{bmatrix} 44 & 6 & 6 \\ 28 & 3 & 21 \end{bmatrix}$