Answer
The translated matrix is $B=\left[ \begin{matrix}
0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
1 & 1 & \frac{3}{2} & \frac{3}{2} & \frac{7}{2} & \frac{7}{2} \\
\end{matrix} \right]$
The resulting graph is shown below:
Work Step by Step
To reduce L to half its perimeter, we will multiply each $x\text{-coordinate}$ and $y\text{-coordinate}$ by $\frac{1}{2}$. That is,
$\begin{align}
& B=\frac{1}{2}\times \left[ \begin{matrix}
0 & 3 & 3 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 & 5 & 5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\
\end{matrix} \right]
\end{align}$
Now, we will add 1 to each $y\text{-coordinate}$ to move the reduced image 1 unit up.
That is,
$\begin{align}
& B=\left[ \begin{matrix}
0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & \frac{1}{2} & \frac{5}{2} & \frac{5}{2} \\
\end{matrix} \right]+\left[ \begin{matrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 & 1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & \frac{3}{2} & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\
1 & 1 & \frac{3}{2} & \frac{3}{2} & \frac{7}{2} & \frac{7}{2} \\
\end{matrix} \right]
\end{align}$
