Answer
$\begin{align}
\left[ \begin{matrix}
0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
2 & 2 & \frac{5}{2} & \frac{5}{2} & \frac{9}{2} & \frac{9}{2} \\
\end{matrix} \right]
\end{align}$
The graph is shown below:
Work Step by Step
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.
