1. **Problem Statement:** Find the rank of the matrix $$\begin{bmatrix} 6 & 1 & 3 & 8 \\ 4 & 2 & 6 & -1 \\ 10 & 3 & 9 & 7 \\ 16 & 4 & 12 & 15 \end{bmatrix}$$ 2. **Recall:** The rank of a matrix is the maximum number of linearly independent rows or columns. We find it by reducing the matrix to row echelon form (REF) or reduced row echelon form (RREF) and counting nonzero rows. 3. **Step 1: Write the matrix** $$A = \begin{bmatrix} 6 & 1 & 3 & 8 \\ 4 & 2 & 6 & -1 \\ 10 & 3 & 9 & 7 \\ 16 & 4 & 12 & 15 \end{bmatrix}$$ 4. **Step 2: Use row operations to get zeros below the first pivot (6 in row 1, column 1):** - Replace $R_2$ by $R_2 - \frac{4}{6} R_1 = R_2 - \frac{2}{3} R_1$ - Replace $R_3$ by $R_3 - \frac{10}{6} R_1 = R_3 - \frac{5}{3} R_1$ - Replace $R_4$ by $R_4 - \frac{16}{6} R_1 = R_4 - \frac{8}{3} R_1$ Calculate each: $$R_2 = \begin{bmatrix}4 & 2 & 6 & -1\end{bmatrix} - \frac{2}{3} \times \begin{bmatrix}6 & 1 & 3 & 8\end{bmatrix} = \begin{bmatrix}4 - 4 & 2 - \frac{2}{3} & 6 - 2 & -1 - \frac{16}{3}\end{bmatrix} = \begin{bmatrix}0 & \frac{4}{3} & 4 & -\frac{19}{3}\end{bmatrix}$$ $$R_3 = \begin{bmatrix}10 & 3 & 9 & 7\end{bmatrix} - \frac{5}{3} \times \begin{bmatrix}6 & 1 & 3 & 8\end{bmatrix} = \begin{bmatrix}10 - 10 & 3 - \frac{5}{3} & 9 - 5 & 7 - \frac{40}{3}\end{bmatrix} = \begin{bmatrix}0 & \frac{4}{3} & 4 & -\frac{19}{3}\end{bmatrix}$$ $$R_4 = \begin{bmatrix}16 & 4 & 12 & 15\end{bmatrix} - \frac{8}{3} \times \begin{bmatrix}6 & 1 & 3 & 8\end{bmatrix} = \begin{bmatrix}16 - 16 & 4 - \frac{8}{3} & 12 - 8 & 15 - \frac{64}{3}\end{bmatrix} = \begin{bmatrix}0 & \frac{4}{3} & 4 & -\frac{19}{3}\end{bmatrix}$$ 5. **Step 3: The matrix now is** $$\begin{bmatrix} 6 & 1 & 3 & 8 \\ 0 & \frac{4}{3} & 4 & -\frac{19}{3} \\ 0 & \frac{4}{3} & 4 & -\frac{19}{3} \\ 0 & \frac{4}{3} & 4 & -\frac{19}{3} \end{bmatrix}$$ 6. **Step 4: Subtract $R_2$ from $R_3$ and $R_4$ to get zeros:** $$R_3 = R_3 - R_2 = \begin{bmatrix}0 & 0 & 0 & 0\end{bmatrix}$$ $$R_4 = R_4 - R_2 = \begin{bmatrix}0 & 0 & 0 & 0\end{bmatrix}$$ 7. **Step 5: The matrix reduces to** $$\begin{bmatrix} 6 & 1 & 3 & 8 \\ 0 & \frac{4}{3} & 4 & -\frac{19}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ 8. **Step 6: Since there are 2 nonzero rows, the rank of the matrix is** $$\boxed{2}$$