## Precalculus (6th Edition) Blitzer

\begin{align} \left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 2 & 2 & \frac{5}{2} & \frac{5}{2} & \frac{9}{2} & \frac{9}{2} \\ \end{matrix} \right] \end{align} The graph is shown below:
To reduce the perimeter of the graph above to half, we will multiply the matrix $B$ by $\frac{1}{2}$ as follows: \begin{align} & \frac{1}{2}B=\frac{1}{2}\left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\ \end{matrix} \right] \end{align} To shift the reduced figure up by 2 units, we will add the following matrix, which represents the reduced figure. $\left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 2 & 2 & 2 & 2 & 2 \\ \end{matrix} \right]$ And the required coordinates to the matrix above and the resultant matrix will be: \begin{align} & \left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 2 & 2 & 2 & 2 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 0+0 & \frac{3}{2}+0 & \frac{3}{2}+0 & \frac{1}{2}+0 & \frac{1}{2}+0 & 0+0 \\ 0+2 & 0+2 & \frac{1}{2}+2 & \frac{1}{2}+2 & \frac{5}{2}+2 & \frac{5}{2}+2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ 2 & 2 & \frac{5}{2} & \frac{5}{2} & \frac{9}{2} & \frac{9}{2} \\ \end{matrix} \right] \end{align} The required coordinates to draw the shifted letter L are as follows: $\left( 0,2 \right),\left( \frac{3}{2},2 \right),\left( \frac{3}{2},\frac{5}{2} \right),\left( \frac{1}{2},\frac{5}{2} \right),\left( \frac{1}{2},\frac{9}{2} \right)$ and $\left( 0,\frac{9}{2} \right)$. Plot the points and trace them to obtain the curve. By subtracting the matrix $\left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 2 & 2 & 2 & 2 & 2 \\ \end{matrix} \right]$ from matrix $\frac{1}{2}B$, and plotting the obtained coordinates, the perimeter of the graph traced was reduced to half and shifted 2 units up from the original.