We are given two matrices, $A = \begin{bmatrix} 1 & 3 & 0 \\ 5 & 4 & 3 \\ 0 & 3 & 2 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 9 & 1 & 0 \end{bmatrix}$. We need to find the product $EA^T$.

1. Problem Description

We are given two matrices, $A = \begin{bmatrix} 1 & 3 & 0 \\ 5 & 4 & 3 \\ 0 & 3 & 2 \end{bmatrix}$ and $E = \begin{bmatrix} 1 & 0 & 3 \\ 9 & 1 & 0 \end{bmatrix}$. We need to find the product $EA^T$.

2. Solution Steps

First, we need to find the transpose of matrix $A$, denoted as $A^T$. The transpose of a matrix is obtained by interchanging its rows and columns. So,

$A^T = \begin{bmatrix} 1 & 5 & 0 \\ 3 & 4 & 3 \\ 0 & 3 & 2 \end{bmatrix}$.

Next, we multiply the matrix $E$ by $A^T$. The matrix $E$ is a $2 \times 3$ matrix, and $A^T$ is a $3 \times 3$ matrix. Therefore, their product will be a $2 \times 3$ matrix.

The elements of the product $EA^T$ are calculated as follows:

$(EA^T)_{11} = (1)(1) + (0)(3) + (3)(0) = 1 + 0 + 0 = 1$

$(EA^T)_{12} = (1)(5) + (0)(4) + (3)(3) = 5 + 0 + 9 = 14$

$(EA^T)_{13} = (1)(0) + (0)(3) + (3)(2) = 0 + 0 + 6 = 6$

$(EA^T)_{21} = (9)(1) + (1)(3) + (0)(0) = 9 + 3 + 0 = 12$

$(EA^T)_{22} = (9)(5) + (1)(4) + (0)(3) = 45 + 4 + 0 = 49$

$(EA^T)_{23} = (9)(0) + (1)(3) + (0)(2) = 0 + 3 + 0 = 3$

Therefore, $EA^T = \begin{bmatrix} 1 & 14 & 6 \\ 12 & 49 & 3 \end{bmatrix}$.

3. Final Answer

$\begin{bmatrix} 1 & 14 & 6 \\ 12 & 49 & 3 \end{bmatrix}$