We are given two matrices, $A$ and $B$, and we need to find the matrix product $BA$. The matrix $A$ is a $3 \times 3$ matrix: $A = \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}$ The matrix $B$ is a $3 \times 3$ matrix: $B = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix}$

1. Problem Description

We are given two matrices, $A$ and $B$, and we need to find the matrix product $BA$.

The matrix $A$ is a $3 \times 3$ matrix:

$A = \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}$

The matrix $B$ is a $3 \times 3$ matrix:

$B = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix}$

2. Solution Steps

To find the matrix product $BA$, we multiply the rows of $B$ by the columns of $A$.

The resulting matrix will be a $3 \times 3$ matrix.

$BA = \begin{bmatrix} 1 & 1 & 4 \\ 8 & 3 & 1 \\ 3 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 7 & 2 & 8 \\ 1 & 0 & 4 \end{bmatrix}$

$BA_{11} = (1)(2) + (1)(7) + (4)(1) = 2 + 7 + 4 = 13$

$BA_{12} = (1)(1) + (1)(2) + (4)(0) = 1 + 2 + 0 = 3$

$BA_{13} = (1)(0) + (1)(8) + (4)(4) = 0 + 8 + 16 = 24$

$BA_{21} = (8)(2) + (3)(7) + (1)(1) = 16 + 21 + 1 = 38$

$BA_{22} = (8)(1) + (3)(2) + (1)(0) = 8 + 6 + 0 = 14$

$BA_{23} = (8)(0) + (3)(8) + (1)(4) = 0 + 24 + 4 = 28$

$BA_{31} = (3)(2) + (2)(7) + (0)(1) = 6 + 14 + 0 = 20$

$BA_{32} = (3)(1) + (2)(2) + (0)(0) = 3 + 4 + 0 = 7$

$BA_{33} = (3)(0) + (2)(8) + (0)(4) = 0 + 16 + 0 = 16$

$BA = \begin{bmatrix} 13 & 3 & 24 \\ 38 & 14 & 28 \\ 20 & 7 & 16 \end{bmatrix}$

3. Final Answer

$\begin{bmatrix} 13 & 3 & 24 \\ 38 & 14 & 28 \\ 20 & 7 & 16 \end{bmatrix}$