Chapter 7.6, Problem 62E

To determine

To solve: The given system of linear equations using inverse of the coefficient matrix.

Answer to Problem 62E

The solution of the system is $I_1=0.7\text{ amperes}$ , $I_2=1.4\text{ amperes}$ and $I_3=2.1\text{ amperes}$ .

Explanation of Solution

Given information :

The circuit:

  

The system of linear equations:

  $\begin{array}{l}\text{2}{I}_1+\text{4}{I}_3=E_1\\ I_2+4I_3=E_2\\ I_1+I_2-I_3=\text{0}\end{array}$

The values of $E_1$ and $E_2$ :

  $E_1=10\text{ volts, }{E}_2=10\text{ volts}$

Formula used :

To solve a system of linear equations:

If $A$ is an invertible matrix, the system of linear equations represented by $AX=B$ has a unique solution given by $X=A^{-1}B$ .

Calculation :

As per the problem,

Put the given values of $E_1$ and $E_2$ in the given system of linear equations:

  $\begin{array}{l}\text{2}{I}_1+\text{4}{I}_3=10\\ I_2+4I_3=10\\ I_1+I_2-I_3=\text{0}\end{array}$

Write the system in the matrix form $AX=B$

  $\left[\begin{array}{ccc}2& 0& 4\\ 0& 1& 4\\ 1& 1& -1\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right]=\left[\begin{array}{c}10\\ 10\\ 0\end{array}\right]$

Where $A=\left[\begin{array}{ccc}2& 0& 4\\ 0& 1& 4\\ 1& 1& -1\end{array}\right]$ , $X=\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right]$ and $B=\left[\begin{array}{c}10\\ 10\\ 0\end{array}\right]$

Then, use Gauss-Jordan elimination to find $A^{-1}$ .

  $A^{-1}=\left[\begin{array}{ccc}0.3571& -0.2857& 0.2857\\ -0.2857& 0.4286& 0.5714\\ 0.0714& 0.1428& -0.1428\end{array}\right]$

Finally, multiply $B$ by $A^{-1}$ on the left to obtain the solution:

  $\begin{array}{l}X=A^{-1}B\\ =\left[\begin{array}{ccc}0.3571& -0.2857& 0.2857\\ -0.2857& 0.4286& 0.5714\\ 0.0714& 0.1428& -0.1428\end{array}\right]\left[\begin{array}{c}10\\ 10\\ 0\end{array}\right]=\left[\begin{array}{c}0.7\\ 1.4\\ 2.1\end{array}\right]\end{array}$

The solution of the system is $I_1=0.7\text{ amperes}$ , $I_2=1.4\text{ amperes}$ and $I_3=2.1\text{ amperes}$ .