Chapter 7, Problem 17CT

To determine

To solve: The given system for the traffic flow.

Answer to Problem 17CT

Solution of the given system is:

  $\begin{array}{l}{x}_1=700-b-a\\ x_2=300-a-b\\ x_3=a\\ x_4=100-b\\ x_5=b\end{array}$

Explanation of Solution

Given information: A figure showing the flow of traffic (in vehicles per hour) through a network of streets ifs given.

Calculation:

Examining the figure,

For upper left, $400+x_2=x_1$

For upper right, $x_1+x_3=x_4+600$

For lower left, $300=x_2+x_3+x_5$

For lower right, $x_5+x_4=100$

The above system of equations can be written as:

  $\begin{array}{l}{x}_1-x_2=400\\ x_1+x_3-x_4=600\\ x_2+x_3+x_5=300\\ x_4+x_5=100\end{array}$

The augmented matrix is:

  $\left[\begin{array}{l}1-1000⋮400\\ 101-10⋮600\\ 01101⋮300\\ 00011⋮100\end{array}\right]$

Operating $R_2\to R_2-R_1,$

  $\left[\begin{array}{l}1-1000⋮400\\ 011-10⋮200\\ 01101⋮300\\ 00011⋮100\end{array}\right]$

Operating $R_3\to R_3-R_2,$

  $\left[\begin{array}{l}1-1000⋮400\\ 011-10⋮200\\ 00011⋮100\\ 00011⋮100\end{array}\right]$

Operating $R_5\to R_5-R_4,$

  $\left[\begin{array}{l}1-1000⋮400\\ 011-10⋮200\\ 00011⋮100\\ 00000⋮0\end{array}\right]$

Operating $R_2\to R_2+R_3,$

  $\left[\begin{array}{l}1-1000⋮400\\ 01101⋮300\\ 00011⋮100\\ 00000⋮0\end{array}\right]$

Operating $R_1\to R_1+R_2,$

  $\left[\begin{array}{l}10101⋮700\\ 01101⋮300\\ 00011⋮100\\ 00000⋮0\end{array}\right]$

The matrix is now in row-echelon form and the corresponding system is:

  $\begin{array}{l}{x}_1+x_3+x_5=700\\ x_2+x_3+x_5=300\\ x_4+x_5=100\end{array}$

Let $x_5=b$

  $\Rightarrow x_4=100-b$

Let $x_3=a$

  $\begin{array}{l}\Rightarrow x_2=300-a-b\\ \&x_1=700-b-a\end{array}$

Thus, solution of the given system is:

  $\begin{array}{l}{x}_1=700-b-a\\ x_2=300-a-b\\ x_3=a\\ x_4=100-b\\ x_5=b\end{array}$