In Exercises 5-8, write a matrix equation that determines the loop currents. [M] If MATLAB or another matrix program is available, solve the system for the loop currents.

In loop 1, current \({I_3}\) is not flowing. Current \({I_1}\) has four \(RI\) voltage drops and \({I_2}\) is negative as it flows in the opposite direction. The voltage drop for \({I_4}\) and \({I_5}\) is negative.

So, the resistance vector for loop 1 is

\({r_1} = \left[ {\begin{array}{*{20}{c}}{15}\\{ - 5}\\0\\{ - 5}\\{ - 1}\end{array}} \right]\).

In loop 2, current \({I_4}\) is not flowing. Current \({I_1}\) has a negative voltage drop and \({I_2}\) has four \(RI\) drops. Voltage drop for \({I_3}\) and \({I_5}\) is negative.

So, theresistance vector for loop 2 is

\({r_2} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{15}\\{ - 5}\\0\\{ - 2}\end{array}} \right]\).

In loop 3, current \({I_1}\) is not flowing. Current \({I_2}\) has a negative voltage drop. \({I_3}\) has four \(RI\) drops. \({I_4}\) and \({I_5}\) have negative voltage drops.

So, the resistance vector for loop 3 is

\({r_3} = \left[ {\begin{array}{*{20}{c}}0\\{ - 5}\\{15}\\{ - 5}\\{ - 3}\end{array}} \right]\).

In loop 4, current \({I_2}\) is not flowing. Currents \({I_1}\) and \({I_3}\) have negative voltage drops. Current \({I_4}\) has four \(RI\) drops and \({I_5}\) has a negative voltage drop.

So, the resistance vector for loop 4 is

\({r_4} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\0\\{ - 5}\\{15}\\{ - 4}\end{array}} \right]\).

In loop 5, current \({I_5}\) has four \(RI\) drops, and all other currents have negative voltage drops.

So, the resistance vector for loop 5 is

\({r_5} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\\{ - 3}\\{ - 4}\\{10}\end{array}} \right]\).

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{{\bf{r}}_1}}&{{{\bf{r}}_2}}&{{{\bf{r}}_3}}&{{{\bf{r}}_4}}&{{{\bf{r}}_5}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\\{{I_5}}\end{array}} \right] = \left[ v \right]\\\left[ {\begin{array}{*{20}{c}}{15}&{ - 5}&0&{ - 5}&{ - 1}\\{ - 5}&{15}&{ - 5}&0&{ - 2}\\0&{ - 5}&{15}&{ - 5}&{ - 3}\\{ - 5}&0&{ - 5}&{15}&{ - 4}\\{ - 1}&{ - 2}&{ - 3}&{ - 4}&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\\{{I_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{40}\\{ - 30}\\{20}\\{ - 10}\\0\end{array}} \right]\end{array}\]

Consider the matrix \(A = \left[ {\begin{array}{*{20}{c}}{15}&{ - 5}&0&{ - 5}&{ - 1}&{40}\\{ - 5}&{15}&{ - 5}&0&{ - 2}&{ - 30}\\0&{ - 5}&{15}&{ - 5}&{ - 3}&{20}\\{ - 5}&0&{ - 5}&{15}&{ - 4}&{ - 10}\\{ - 1}&{ - 2}&{ - 3}&{ - 4}&{10}&0\end{array}} \right]\).

Use the code in MATLAB to obtain the row-reduced echelon form, as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {\begin{array}{*{20}{c}}{15}&{ - 5}&0&{ - 5}&{ - 1}&{ - 40}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 5}&{15}&{ - 5}&0&{ - 1}&{30}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&{ - 5}&{15}&{ - 5}&{ - 3}&{ - 20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 5}&0&{ - 5}&{15}&{ - 4}&{10;\,\,\begin{array}{*{20}{c}}{ - 1}&{ - 2}&{ - 3}&{ - 4}&{10}&0\end{array}}\end{array}{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}{15}&{ - 5}&0&{ - 5}&{ - 1}&{40}\\{ - 5}&{15}&{ - 5}&0&{ - 2}&{ - 30}\\0&{ - 5}&{15}&{ - 5}&{ - 3}&{20}\\{ - 5}&0&{ - 5}&{15}&{ - 4}&{ - 10}\\{ - 1}&{ - 2}&{ - 3}&{ - 4}&{10}&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&0&{ - 3.37}\\0&1&0&0&0&{ - 0.11}\\0&0&1&0&0&{ - 2.27}\\0&0&0&1&0&{ - 1.67}\\0&0&0&0&1&{ - 1.70}\end{array}} \right]\)

Here, the negative sign represents the direction of the current in the loop.

\(\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\\{{I_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3.37}\\{0.11}\\{2.27}\\{1.67}\\{1.70}\end{array}} \right]\)

So, the loop currents in the given circuit are \(\left[ {\begin{array}{*{20}{c}}{3.37}\\{0.11}\\{2.27}\\{1.67}\\{1.70}\end{array}} \right]\).