1. **State the problem:** We need to find the element in the 3rd row and 2nd column of the product matrix $AB$, denoted as $(AB)_{3,2}$. 2. **Recall matrix multiplication rule:** The element $(AB)_{i,j}$ is calculated as the dot product of the $i$th row of $A$ and the $j$th column of $B$: $$ (AB)_{i,j} = \sum_{k=1}^n A_{i,k} B_{k,j} $$ where $n$ is the number of columns in $A$ (which must equal the number of rows in $B$). 3. **Check dimensions:** Matrix $A$ is $3 \times 4$ and matrix $B$ is $4 \times 2$, so multiplication $AB$ is defined and the result is a $3 \times 2$ matrix. 4. **Extract the 3rd row of $A$ and 2nd column of $B$:** $$ A_{3,:} = [-1, -2, -3, -4] $$ $$ B_{:,2} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ -1 \end{bmatrix} $$ 5. **Calculate $(AB)_{3,2}$:** $$ (AB)_{3,2} = (-1) \times 0 + (-2) \times 1 + (-3) \times 0 + (-4) \times (-1) $$ $$ = 0 - 2 + 0 + 4 $$ $$ = 2 $$ 6. **Final answer:** $(AB)_{3,2} = 2$. Therefore, the correct choice is E. 2.