## Precalculus (6th Edition) Blitzer

To shift the figure left by 2 units and down by 3 units, we will subtract the matrix $B$ by the matrix below: $\left[ \begin{matrix} 2 & 2 & 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 & 3 & 3 \\ \end{matrix} \right]$ And the required coordinates will be obtained as below: \begin{align} & \left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right]-\left[ \begin{matrix} 2 & 2 & 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 & 3 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 0-2 & 3-2 & 3-2 & 1-2 & 1-2 & 0-2 \\ 0-3 & 0-3 & 1-3 & 1-3 & 5-3 & 5-3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2 & 1 & 1 & -1 & -1 & -2 \\ -3 & -3 & -2 & -2 & 2 & 2 \\ \end{matrix} \right] \end{align} The required coordinates to draw the shifted letter L are as follows: $\left( -2,-3 \right),\left( 1,-3 \right),\left( 1,-2 \right),\left( -1,-2 \right),\left( -1,2 \right)$ and $\left( -2,2 \right)$. Plot the points and trace them to obtain the curve. By subtracting the matrix $\left[ \begin{matrix} 2 & 2 & 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 & 3 & 3 \\ \end{matrix} \right]$ from matrix B, and plotting the obtained coordinates, the graph traced was shifted 2 units left and 3 units down from the original.