## Thinking Mathematically (6th Edition)

(a) Find $\left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]\times \left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]$as, \begin{align} & \left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]\times \left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} 2\cdot 0+3\cdot 5 & 2\cdot 1+3\cdot 6 \\ 4\cdot 0+7\cdot 5 & 4\cdot 1+7\cdot 6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0+15 & 2+18 \\ 0+35 & 4+42 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 15 & 20 \\ 35 & 46 \\ \end{matrix} \right] \end{align} Hence, $\left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]\times \left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} 15 & 20 \\ 35 & 46 \\ \end{matrix} \right]$. (b) Find $\left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]$as, \begin{align} & \left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 0\cdot 2+1\cdot 4 & 0\cdot 3+1\cdot 7 \\ 5\cdot 2+6\cdot 4 & 5\cdot 3+6\cdot 7 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0+4 & 0+7 \\ 10+24 & 15+42 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4 & 7 \\ 34 & 57 \\ \end{matrix} \right] \end{align} Hence, $\left[ \begin{matrix} 0 & 1 \\ 5 & 6 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 3 \\ 4 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 7 \\ 34 & 57 \\ \end{matrix} \right]$. (c) The row of first set is multiplied with the column of second set. Thus, the row of first is distributed over column of second. Hence, distributive property is used in above multiplication.