buy Now compute $ B = A_2 A_1 $: $ A_1 $ columns are [0, 1] and [0, 0]. $ A_2 $ rows are [0, 0] and [1, 0]. $ B_{11} = [0, 0] \cdot [0, 0] = 0 $, $ B_{12} = [0, 0] \cdot [0, 0] = 0 $. $ B_{21} = [1, 0] \cdot [0, 0] = 0 $, $ B_{22} = [1, 0] \cdot [0, 0] = 0 $. So, $ B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $, and the first row is all 0s. However, if we adjust to your intent with $ A_1 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $ A_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $: $ B = A_2 A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $, and the first row is [0, 1], which is not all 0s. This counterexample shows that if $ A_1 $'s first row is all 0s but the preceding matrices (like $ A_2 $) can introduce non-zero elements, the first row of $ B $ can be non-zero. Since the statement claims this holds "in any defined multiplication," and we've found a