Chapter 8.3, Problem 65E

To determine

The unknown currents for the given voltages by using the inverse of the coefficient matrix.

Answer to Problem 65E

  $x=0.5,y=3,$ and $z=3.5$ .

Explanation of Solution

Given information:

The currents $I_1,I_2,$ and $I_3$ , in amperes, are the solution of the system of linear equation.

  $\{\begin{array}{l}2I_1+4I_3=E_1\\ I_2+4I_3=E_2\\ I_1+I_2-I_3=0\end{array}$

Where $E_1$ and $E_2$ are voltages. Use the inverse of the coefficient matrix of this system to find the unknown currents for the given voltages.

  

  $\begin{array}{l}{E}_1=15volts\\ E_2=17volts\end{array}$

Calculation:

The given system of equation is:

  $\{\begin{array}{l}2I_1+4I_3=E_1\\ I_2+4I_3=E_2\\ I_1+I_2-I_3=0\end{array}\Rightarrow \{\begin{array}{l}2I_1+4I_3=15\\ I_2+4I_3=17\\ I_1+I_2-I_3=0\end{array}$

Convert the augmented matrix into the row-echelon form by gauss elimination method:

Firstly find the pivot in the first column and swap the third and the first row:

  $\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 4\\ 2& 0& 4\end{array}|\begin{array}{c}0\\ 17\\ 15\end{array}\right]$

Eliminate the first column and find the pivot in the second column in the second row :

  $\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 4\\ 0& -2& 6\end{array}|\begin{array}{c}0\\ 17\\ 15\end{array}\right]$

Eliminate the second column and make the pivot in the third column by dividing the third row by $14$ :

  $\left[\begin{array}{ccc}1& 0& -5\\ 0& 1& 4\\ 0& 0& 1\end{array}|\begin{array}{c}-17\\ 17\\ 3.5\end{array}\right]$

Eliminate the third column:

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{c}0.5\\ 3\\ 3.5\end{array}\right]$

Therefore, the solutions are $x=0.5,y=3,$ and $z=3.5$ .