1. **State the problem:** Find the rank of the matrix $$\begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 3 & 2 & 2 \\ 2 & 4 & 3 & 4 \\ 3 & 7 & 5 & 6 \end{bmatrix}$$ by reducing it to its normal (row echelon) form. 2. **Recall the rank definition:** The rank of a matrix is the number of nonzero rows in its row echelon form. 3. **Start row reduction:** - Subtract row 1 from row 2: $$R_2 \to R_2 - R_1 = (1-1, 3-2, 2-1, 2-2) = (0,1,1,0)$$ - Subtract 2 times row 1 from row 3: $$R_3 \to R_3 - 2R_1 = (2-2, 4-4, 3-2, 4-4) = (0,0,1,0)$$ - Subtract 3 times row 1 from row 4: $$R_4 \to R_4 - 3R_1 = (3-3, 7-6, 5-3, 6-6) = (0,1,2,0)$$ Matrix now: $$\begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \end{bmatrix}$$ 4. **Eliminate below and above pivots:** - Subtract row 2 from row 4: $$R_4 \to R_4 - R_2 = (0,1-1,2-1,0-0) = (0,0,1,0)$$ - Subtract row 3 from row 4: $$R_4 \to R_4 - R_3 = (0,0,1-1,0-0) = (0,0,0,0)$$ Matrix now: $$\begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ 5. **Count nonzero rows:** There are 3 nonzero rows. **Final answer:** The rank of the matrix is $3$.