Chapter 12.2, Problem 10E

Solution

a.

To graph: The directed graph indicating road travel between the houses given that Alice and Becky live on Parkway East, at the intersections of Owens Bridge and Bay Bridge respectively. Carl and David live on Parkway west at the intersection of Bay bridge and Owens Bridge respectively whereas Parkway east is a one-way street running east, Parkway west is one-way running west and both the bridges are two ways.

Given information:

Alice and Becky live that’s why directed with arrows on both up and right. e on Parkway East, at the intersections of Owens Bridge and Bay Bridge respectively. Carl and David live on Parkway west at the intersection of Bay bridge and Owens Bridge respectively. Parkway east is a one-way street running east. Parkway west is one-way running west. Both bridges are two ways.

Graph :

The directed graph will be shown by the arrows representing the direction of the houses facing each other.

Given that Alice $A$ and Becky $B$ live on Parkway East, which is one way headed to East, then the arrow points on the right on the directed graph.

As Cate $C$ and David $B$ live on Parkway west, which is one way headed to west, then the arrow point on the left on the directed graph.

As the bridges are the connectors which are both ways, and that’s why directed with arrows on both up and right direction.

Draw a directed graph indicating the road travel between the houses.

  

Interpretation:

The graph shows the road travelled between the four houses.

b.

To determine : The matrix $T$ to represent the information in the directed graph.

The matrix $T$ is $T=\begin{array}{c}\\ A\\ B\\ C\\ D\end{array}\left[\begin{array}{cccc}A& B& C& D\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 0& 0\end{array}\right]$ .

Given information :

Alice and Becky live that’s why directed with arrows on both up and right. e on Parkway East, at the intersections of Owens Bridge and Bay Bridge respectively. Carl and David live on Parkway west at the intersection of Bay bridge and Owens Bridge respectively. Parkway east is a one-way street running east. Parkway west is one-way running west. Both bridges are two ways.

Explanation :

Consider the directed graph as follows.

  

To write the matrix $T$ , put 1 for the houses which are connected to each other directly, and 0 where the houses are not connected to each other directly.

  $T=\begin{array}{c}\\ A\\ B\\ C\\ D\end{array}\left[\begin{array}{cccc}A& B& C& D\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 0& 0\end{array}\right]$

c.

To calculate : The matrix $T^2$ and explain what does the matrix model.

The matrix $T^2$ is $T^2=\begin{array}{c}\\ A\\ B\\ C\\ D\end{array}\left[\begin{array}{cccc}A& B& C& D\\ 1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 1& 0\\ 0& 1& 0& 1\end{array}\right]$ .

Given information :

Alice and Becky live that’s why directed with arrows on both up and right. e on Parkway East, at the intersections of Owens Bridge and Bay Bridge respectively. Carl and David live on Parkway west at the intersection of Bay bridge and Owens Bridge respectively. Parkway east is a one-way street running east. Parkway west is one-way running west. Both bridges are two ways.

Formula Used:

Use the matrix multiplication.

Explanation :

Consider the matrix $T=\begin{array}{c}\\ A\\ B\\ C\\ D\end{array}\left[\begin{array}{cccc}A& B& C& D\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 0& 0\end{array}\right]$ .

Apply the matrix multiplication to calculate the value of $T^2$ .

  $\begin{array}{c}{T}^2=\left[\begin{array}{cccc}0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 0& 0\end{array}\right]\left[\begin{array}{cccc}0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 0& 0\end{array}\right]\\ =\left[\begin{array}{cccc}0+0+0+1& 0+0+0+0& 0+1+0+0& 0+0+0+0\\ 0+0+0+0& 0+0+1+0& 0+0+0+0& 0+0+1+0\\ 0+0+0+1& 0+0+0+0& 0+1+0+0& 0+0+0+0\\ 0+0+0+0& 1+0+0+0& 0+0+0+0& 1+0+0+0\end{array}\right]\\ =\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 1& 0\\ 0& 1& 0& 1\end{array}\right]\end{array}$

The value of $T^2$ is $T^2=\begin{array}{c}\\ A\\ B\\ C\\ D\end{array}\left[\begin{array}{cccc}A& B& C& D\\ 1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& 1& 0\\ 0& 1& 0& 1\end{array}\right]$