Solution

Elementary Linear algebra

Chapter 1.9

Exercise 5

$\text{in the same direction as the current}$. Hence there is a voltage drop of $E=-2I_2$ V. Then we go from right to left across the bottom of the left loop and we encounter the $6$ V battery $\textit{going from the positive to negative terminal}$. Hence there is a voltage drop of $-6$ V. And finally we come back up the left side and encounter the final $2\ \Omega$ resistor. This time we're going $\textit{in the opposite direction as the current}$ and hence there is a voltage rise of $E=2I_1$ V. Thus, by summing all of the voltage drops and rises we obtain the equation

$$$$-2I_2 - 6+2I_1 = 0

Using exactly the same procedure on the right loop we experience a voltage drop of $-8$ V across the battery, then a voltage rise of $4I_3$ V across the right resistor and a voltage rise of $2I_2$ V across the left resistor. So we obtain the equation

$$$$-8 + 4I_3 + 2I_2 = 0

Answer