1. **Problem:** Determine the rank of the matrix $$A = \begin{pmatrix} 0 & 0 & 2 & 2 & 0 \\ 1 & 3 & 2 & 4 & 1 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{pmatrix}$$ 2. **Formula and rules:** The rank of a matrix is the maximum number of linearly independent rows or columns. We use row reduction (Gaussian elimination) to find the number of nonzero rows in the row echelon form. 3. **Step-by-step row reduction:** Start with $$\begin{pmatrix} 0 & 0 & 2 & 2 & 0 \\ 1 & 3 & 2 & 4 & 1 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{pmatrix}$$ Swap row 1 and row 2 to get a leading 1: $$\begin{pmatrix} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{pmatrix}$$ Eliminate below row 1: - Row 3 = Row 3 - 2*Row 1: $$\begin{pmatrix} 2 & 6 & 2 & 6 & 2 \end{pmatrix} - 2 \times \begin{pmatrix} 1 & 3 & 2 & 4 & 1 \end{pmatrix} = \begin{pmatrix} 2-2 & 6-6 & 2-4 & 6-8 & 2-2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -2 & -2 & 0 \end{pmatrix}$$ - Row 4 = Row 4 - 3*Row 1: $$\begin{pmatrix} 3 & 9 & 1 & 10 & 6 \end{pmatrix} - 3 \times \begin{pmatrix} 1 & 3 & 2 & 4 & 1 \end{pmatrix} = \begin{pmatrix} 3-3 & 9-9 & 1-6 & 10-12 & 6-3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -5 & -2 & 3 \end{pmatrix}$$ Matrix now: $$\begin{pmatrix} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & -2 & -2 & 0 \\ 0 & 0 & -5 & -2 & 3 \end{pmatrix}$$ Add Row 2 and Row 3: $$\begin{pmatrix} 0 & 0 & 2 & 2 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & -2 & -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ So Row 3 is dependent on Row 2. Check Row 4: Try to express Row 4 as a combination of Row 2: Multiply Row 2 by $-\frac{5}{2}$: $$-\frac{5}{2} \times \begin{pmatrix} 0 & 0 & 2 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -5 & -5 & 0 \end{pmatrix}$$ Row 4 is $$\begin{pmatrix} 0 & 0 & -5 & -2 & 3 \end{pmatrix}$$ which is not equal to the above, so Row 4 is independent. 4. **Conclusion:** Nonzero rows after reduction are Row 1, Row 2, and Row 4. Therefore, $$\text{rank}(A) = 3$$.