Answer
See solution
Work Step by Step
$\mathrm{r}_{1}=\left[\begin{array}{c}12 \\ -7 \\ 0 \\ -4\end{array}\right]$ Voltagedrop for I negative; Is flows in opposite direction
Loop 2
$\mathbf{r}_{2}=\left[\begin{array}{c}-7 \\ 15 \\ -6 \\ 0\end{array}\right] \begin{array}{c}\text { Voltage drop for } \mathrm{I}_{1} \text { is negative; } \mathrm{I}_{1} \text { flows in opposite direction } \\ \text { Total of three RI voltage drops for current } I_{2} \\ \text { Voltage drop for I is negative; I }_{3} \text { flows in opposite direction } \\ \text { Current I }_{4} \text { does not flow in loop } 2\end{array}$
\[
r_{3}=\left[\begin{array}{c}
0 \\
-6 \\
14 \\
-5
\end{array}\right], r_{4}=\left[\begin{array}{c}
-4 \\
0 \\
-5 \\
13
\end{array}\right], \text { and } R=\left[\begin{array}{cccc}
\mathbf{r}_{1} & \mathbf{r}_{2} & \mathbf{r}_{3} & \mathbf{r}_{4}
\end{array}\right]=\left[\begin{array}{cccc}
12 & -7 & 0 & -4 \\
-7 & 15 & -6 & 0 \\
0 & -6 & 14 & -5 \\
-4 & 0 & -5 & 13
\end{array}\right]
\]
Next, set
\[
\mathbf{v}=\left[\begin{array}{c}
40 \\
30 \\
20 \\
-10
\end{array}\right]
\]
See the negative voltage in loop 4. The current direction chosen in loop 4 is opposed by the orientation of the voltage source in that loop. So Ri $=\mathrm{v}$ becomes
\[
\left[\begin{array}{cccc}
12 & -7 & 0 & -4 \\
-7 & 15 & -6 & 0 \\
0 & -6 & 14 & -5 \\
-4 & 0 & -5 & 13
\end{array}\right]\left[\begin{array}{c}
I_{1} \\
I_{2} \\
I_{3} \\
I_{4}
\end{array}\right]=\left[\begin{array}{c}
40 \\
30 \\
20 \\
-10
\end{array}\right]
\]
Solution
\[
i=\left[\begin{array}{c}
I_{1} \\
I_{2} \\
I_{3} \\
I_{4}
\end{array}\right]=\left[\begin{array}{c}
11.43 \\
10.55 \\
8.04 \\
5.84
\end{array}\right]
\]
