Solution
= Solution
Using row-column <matrix multiplication>, the second row of the first <matrix> is $(1,6)$. Therefore
$$
\boxed{\begin{pmatrix}3&7\\1&6\end{pmatrix}\begin{pmatrix}2&4\\0&0\end{pmatrix}=\begin{pmatrix}6&12\\2&4\end{pmatrix}.}
$$
= Solution
Using row-column <matrix multiplication>, the second row of the first <matrix> is $(1,6)$. Therefore
$$
\boxed{\begin{pmatrix}3&7\\1&6\end{pmatrix}\begin{pmatrix}2&4\\0&0\end{pmatrix}=\begin{pmatrix}6&12\\2&4\end{pmatrix}.}
$$