91Ó°ÊÓ

The given system of the equations are

$\left\{\begin{array}{l}\begin{array}{c}-4+8-2I_2=0\\ 8=5I_4+I_1\\ 4=3I_3+I_1\end{array}\\ I_3+I_4=I_1\end{array}\right.$

We have to find the currents$I_1,I_2,I_3,\mathrm{and}{I}_4.$

By rewriting the system we get,

$\left\{\begin{array}{l}\begin{array}{c}{I}_2=2\\ I_1+5I_4=8\\ I_1+3I_3=4\end{array}\\ I_1-I_3-I_4=0\end{array}\right.$

Let's write the augmented matrix that represents the system

$$\begin{gathered} \left[\begin{array}{cccccc}0& 1& 0& 0& |& 2\\ 1& 0& 0& 5& |& 8\\ 1& 0& 3& 0& |& 4\\ -1& 0& 1& 1& |& 0\end{array}\right] \\ {R}_1=r_1+r_2 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 1& 0& 0& 5& |& 8\\ 1& 0& 3& 0& |& 4\\ -1& 0& 1& 1& |& 0\end{array}\right] \\ {R}_2=r_2-r_1 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& -1& 0& 0& |& -2\\ 1& 0& 3& 0& |& 4\\ -1& 0& 1& 1& |& 0\end{array}\right] \\ {R}_3=r_3-r_1 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& -1& 0& 0& |& -2\\ 0& -1& 3& -5& |& -6\\ -1& 0& 1& 1& |& 0\end{array}\right] \end{gathered}$$

By proceeding with the calculation further

$$\begin{gathered} R_4=r_4+r_1 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& -1& 0& 0& |& -2\\ 0& -1& 3& -5& |& -6\\ 0& 1& 1& 6& |& 10\end{array}\right] \\ {R}_2=\left(-1\right)r_2 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& -1& 3& -5& |& -6\\ 0& 1& 1& 6& |& 10\end{array}\right] \\ {R}_3=r_2+r_3 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& 0& 3& -5& |& -4\\ 0& 1& 1& 6& |& 10\end{array}\right] \\ {R}_4=r_2-r_4 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& 0& 3& -5& |& -4\\ 0& 0& -1& -6& |& -8\end{array}\right] \end{gathered}$$

By proceeding with the calculation further

$$\begin{gathered} R_3=\frac{r_3}{3} \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 0& -1& -6& |& -8\end{array}\right] \\ {R}_4=r_3+r_4 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 0& 0& \raisebox{1ex}{$-23$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$-28$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right] \\ {R}_4=-\frac{3}{23}{r}_4 \\ \left[\begin{array}{cccccc}1& 1& 0& 5& |& 10\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 0& 0& 1& |& \raisebox{1ex}{$28$}\!\left/ \!\raisebox{-1ex}{$23$}\right.\end{array}\right] \end{gathered}$$

The matrix is now in row echelon form. The final matrix represents the system

$\left\{\begin{array}{l}\begin{array}{c}{I}_1+I_2+5I_4=10\left(1\right)\\ I_2=2\left(2\right)\\ I_3-\frac{5}{3}{I}_4=-\frac{4}{3}\left(3\right)\end{array}\\ I_4=\frac{28}{23}\left(4\right)\end{array}\right.$

To find $I_3$we will substitute $I_4$in the equation in (3)

$$\begin{gathered} I_3-\frac{5}{3}\left(\frac{28}{23}\right)=-\frac{4}{3} \\ {I}_3=-\frac{4}{3}+\frac{140}{69} \\ {I}_3=\frac{16}{23} \end{gathered}$$

To find $I_1$we will substitute $I_2\mathrm{and}{I}_4$in the equation in (1)

$$\begin{gathered} I_1+\left(2\right)+5\left(\frac{28}{23}\right)=10 \\ {I}_1=10-2-\frac{140}{23} \\ {I}_1=\frac{44}{23} \end{gathered}$$

Therefore, the value of currents are$I_1=\frac{44}{23},I_2=2,I_3=\frac{16}{23},\mathrm{and}{I}_4=\frac{28}{23}.$