1. **Stating the problem:** We need to find the product of the column vector $$\begin{bmatrix}0 \\ -2\end{bmatrix}$$ and the row vector $$\begin{bmatrix}-1 & 0 & -1\end{bmatrix}$$. 2. **Formula used:** The product of a column vector of size $$m \times 1$$ and a row vector of size $$1 \times n$$ results in an $$m \times n$$ matrix. Each element of the resulting matrix is the product of the corresponding elements from the column and row vectors. 3. **Calculation:** Let the column vector be $$\mathbf{u} = \begin{bmatrix}u_1 \\ u_2\end{bmatrix} = \begin{bmatrix}0 \\ -2\end{bmatrix}$$ and the row vector be $$\mathbf{v} = \begin{bmatrix}v_1 & v_2 & v_3\end{bmatrix} = \begin{bmatrix}-1 & 0 & -1\end{bmatrix}$$. The product $$\mathbf{u} \cdot \mathbf{v}$$ is: $$ \begin{bmatrix} u_1 v_1 & u_1 v_2 & u_1 v_3 \\ u_2 v_1 & u_2 v_2 & u_2 v_3 \end{bmatrix} = \begin{bmatrix}0 \times (-1) & 0 \times 0 & 0 \times (-1) \\ -2 \times (-1) & -2 \times 0 & -2 \times (-1)\end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & 2\end{bmatrix} $$ 4. **Explanation:** Each element in the resulting matrix is computed by multiplying the corresponding element from the column vector by the corresponding element from the row vector. For example, the element in the second row, first column is $$-2 \times -1 = 2$$. 5. **Final answer:** $$ \begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & 2\end{bmatrix} $$ This matches option A.