The problem asks to find the product of matrix $A$ and identity matrix $I$, i.e., $AI$. The given matrices are $A = \begin{bmatrix} 5 & 7 & 8 \\ 2 & 3 & 1 \\ 1 & -7 & -4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

1. Problem Description

The problem asks to find the product of matrix $A$ and identity matrix $I$, i.e., $AI$. The given matrices are $A = \begin{bmatrix} 5 & 7 & 8 \\ 2 & 3 & 1 \\ 1 & -7 & -4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

2. Solution Steps

To find the product of two matrices, we multiply the rows of the first matrix by the columns of the second matrix. In this case, we are multiplying a $3 \times 3$ matrix $A$ by a $3 \times 3$ identity matrix $I$.

The product $AI$ is given by:

$AI = \begin{bmatrix} 5 & 7 & 8 \\ 2 & 3 & 1 \\ 1 & -7 & -4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

The first element of the resulting matrix is $(5 \times 1) + (7 \times 0) + (8 \times 0) = 5$.

The second element of the first row is $(5 \times 0) + (7 \times 1) + (8 \times 0) = 7$.

The third element of the first row is $(5 \times 0) + (7 \times 0) + (8 \times 1) = 8$.

The first element of the second row is $(2 \times 1) + (3 \times 0) + (1 \times 0) = 2$.

The second element of the second row is $(2 \times 0) + (3 \times 1) + (1 \times 0) = 3$.

The third element of the second row is $(2 \times 0) + (3 \times 0) + (1 \times 1) = 1$.

The first element of the third row is $(1 \times 1) + (-7 \times 0) + (-4 \times 0) = 1$.

The second element of the third row is $(1 \times 0) + (-7 \times 1) + (-4 \times 0) = -7$.

The third element of the third row is $(1 \times 0) + (-7 \times 0) + (-4 \times 1) = -4$.

Therefore, the resulting matrix $AI$ is:

$AI = \begin{bmatrix} 5 & 7 & 8 \\ 2 & 3 & 1 \\ 1 & -7 & -4 \end{bmatrix}$

The product of any matrix and the identity matrix is the matrix itself.

3. Final Answer

$AI = \begin{bmatrix} 5 & 7 & 8 \\ 2 & 3 & 1 \\ 1 & -7 & -4 \end{bmatrix}$