We are given two matrices, $D = \begin{bmatrix} 2 & 6 \\ 7 & 3 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 7 & 1 & 0 \end{bmatrix}$. We are asked to find the product $DE$.

1. Problem Description

We are given two matrices, $D = \begin{bmatrix} 2 & 6 \\ 7 & 3 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 7 & 1 & 0 \end{bmatrix}$. We are asked to find the product $DE$.

2. Solution Steps

To find the product of two matrices $D$ and $E$, we need to check if the number of columns of $D$ is equal to the number of rows of $E$. Here, $D$ is a $2 \times 2$ matrix and $E$ is a $2 \times 3$ matrix. Since the number of columns of $D$ (which is 2) is equal to the number of rows of $E$ (which is 2), we can multiply the matrices. The resulting matrix $DE$ will have the same number of rows as $D$ and the same number of columns as $E$, so $DE$ will be a $2 \times 3$ matrix.

The elements of the product matrix $DE$ are calculated as follows:

$(DE)_{ij} = \sum_{k=1}^{n} D_{ik} E_{kj}$, where $n$ is the number of columns in $D$ (or the number of rows in $E$).

$DE = \begin{bmatrix} 2 & 6 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 3 \\ 7 & 1 & 0 \end{bmatrix} = \begin{bmatrix} (2)(1) + (6)(7) & (2)(0) + (6)(1) & (2)(3) + (6)(0) \\ (7)(1) + (3)(7) & (7)(0) + (3)(1) & (7)(3) + (3)(0) \end{bmatrix}$

$DE = \begin{bmatrix} 2 + 42 & 0 + 6 & 6 + 0 \\ 7 + 21 & 0 + 3 & 21 + 0 \end{bmatrix} = \begin{bmatrix} 44 & 6 & 6 \\ 28 & 3 & 21 \end{bmatrix}$

3. Final Answer

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