To solve this, let's follow the row operation given:

1. Matrix and Row Operation:

   • Original matrix: 1 & 6 & 6 \\ 6 & 8 & -5 \\ -7 & 5 & -8 \end{bmatrix}$$
   • The row operation to perform is: Replace $R_2$ by $-6 \cdot R_1 + R_2$.

2. Applying $-6 \cdot R_1$:

   • $-6 \cdot R_1 = \begin{bmatrix} -6 & -36 & -36 \end{bmatrix}$.

3. Adding $-6 \cdot R_1$ to $R_2$:

   • $R_2 = \begin{bmatrix} 6 & 8 & -5 \end{bmatrix}$
   • Compute each element in $R_2$ by adding the corresponding element in $-6 \cdot R_1$:
     • First element: $6 + (-6) = 0$
     • Second element: $8 + (-36) = -28$
     • Third element: $-5 + (-36) = -41$

   So, the new $R_2$ row becomes $\begin{bmatrix} 0 & -28 & -41 \end{bmatrix}$.

4. Resulting Matrix:

   • The matrix after performing the row operation is: 1 & 6 & 6 \\ 0 & -28 & -41 \\ -7 & 5 & -8 \end{bmatrix}$$

   Now, you can fill in the missing values in the answer boxes as follows:

   • Second row: $0, -28, -41$
   • Third row remains: $-7, 5, -8$

Would you like a further breakdown or have any questions?

Here are 5 related questions that may help you deepen your understanding:

1. What would happen if we used a different scalar in front of $R_1$ for this row operation?
2. How do row operations affect the determinant of a matrix?
3. Why is the identity matrix unaffected by similar row operations?
4. What is the significance of achieving zeros in matrix row operations?
5. How do row operations contribute to finding the inverse of a matrix?

Tip: When performing row operations, always carefully track signs and addition/subtraction to avoid errors.