Chapter 7.4, Problem 89E

To determine

To calculate:

To find the values of the amperes $I_1,I_2$ and $I_3$ .

Answer to Problem 89E

The values of the amperes are

  $\begin{array}{l}{I}_1=1.3\\ I_2=2.2\\ I_3=0.9\end{array}$

Explanation of Solution

Given information:

The given currents in the electrical networks are

  $\begin{array}{l}{I}_1-I_2+I_3=0\\ 2I_1+2I_2=7\\ 2I_2+4I_3=8\end{array}$

Calculation:

The given currents in the electrical networks are

  $\begin{array}{l}{I}_1-I_2+I_3=0\to \left(1\right)\\ 2I_1+2I_2=7\to \left(2\right)\\ 2I_2+4I_3=8\to \left(3\right)\end{array}$

The given system of equations can be written in matrix notation as shown below,

  $\left(\begin{array}{ccc}1& -1& 1\\ 2& 2& 0\\ 0& 2& 4\end{array}\right)\left(\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right)=\left(\begin{array}{c}0\\ 7\\ 8\end{array}\right)$

Multiply both sides by the inverse of the coefficient matrix.

Note that the product of a matrix with its inverse is an identity matrix of the same order.

  $\left(\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right)={\left(\begin{array}{ccc}1& -1& 1\\ 2& 2& 0\\ 0& 2& 4\end{array}\right)}^{-1}\left(\begin{array}{c}0\\ 7\\ 8\end{array}\right)$

Consider, ${\left(\begin{array}{ccc}1& -1& 1\\ 2& 2& 0\\ 0& 2& 4\end{array}\right)}^{-1}$

Augment with a $3\times 3$ identity matrix,

  $=\left[\begin{array}{ccccccc}1& -1& 1& \mid & 1& 0& 0\\ 2& 2& 0& \mid & 0& 1& 0\\ 0& 2& 4& \mid & 0& 0& 1\end{array}\right]$

Reduce the matrix to row echelon form,

  $=\left[\begin{array}{ccccccc}2& 2& 0& \mid & 0& 1& 0\\ 0& -2& 1& \mid & 1& -\frac{1}{2}& 0\\ 0& 0& 5& \mid & 1& -\frac{1}{2}& 1\end{array}\right]$

Reduce further,

  $=\left[\begin{array}{ccccccc}1& 0& 0& \mid & \frac{2}{5}& \frac{3}{10}& -\frac{1}{10}\\ 0& 1& 0& \mid & -\frac{2}{5}& \frac{1}{5}& \frac{1}{10}\\ 0& 0& 1& \mid & \frac{1}{5}& -\frac{1}{10}& \frac{1}{5}\end{array}\right]$

Hence, we get

  $=\left(\begin{array}{ccc}\frac{2}{5}& \frac{3}{10}& -\frac{1}{10}\\ -\frac{2}{5}& \frac{1}{5}& \frac{1}{10}\\ \frac{1}{5}& -\frac{1}{10}& \frac{1}{5}\end{array}\right)$

Now, $\left(\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right)={\left(\begin{array}{ccc}1& -1& 1\\ 2& 2& 0\\ 0& 2& 4\end{array}\right)}^{-1}\left(\begin{array}{c}0\\ 7\\ 8\end{array}\right)$

  $=\left(\begin{array}{ccc}\frac{2}{5}& \frac{3}{10}& -\frac{1}{10}\\ -\frac{2}{5}& \frac{1}{5}& \frac{1}{10}\\ \frac{1}{5}& -\frac{1}{10}& \frac{1}{5}\end{array}\right)\left(\begin{array}{c}0\\ 7\\ 8\end{array}\right)$

  $=\left(\begin{array}{c}\frac{13}{10}\\ \frac{11}{5}\\ \frac{9}{10}\end{array}\right)$

Therefore,

  $\begin{array}{l}{I}_1=1.3\\ I_2=2.2\\ I_3=0.9\end{array}$

Hence, The given currents in the electrical networks are

  $\begin{array}{l}{I}_1-I_2+I_3=0\\ 2I_1+2I_2=7\\ 2I_2+4I_3=8\end{array}$