Another example of a linear system is a circuit with N power supplies and N resistors in parallel. Let Rj denote the resistance of each resistor and Ej denote the electromotive force (or EMF) generated by each power supply. Suppose we would like to know the potential difference VV0 between two points on opposite sides of the circuit, and the N currents Ij (shown in the diagram below) at each branch of this circuit. We can assume that V0=0 and replace VV0 with V always since only potential difference matters. Using Ohm's Law for each branch, we obtain N equations IjRj+V=Ej. for j=0,,N1 We also obtain one more equation using Kirchoff's Law: j=0N1Ij=0 You want to find the vector I=I0I1IN1V containing the N currents at each branch, and the potential difference between the two sides. That is, we want to solve the linear system []I0I1IN1V=E0E1EN10 or AI=E~ given by the above N+1 equations. The first thing you will need to do is determine the matrix A so that the matrix-vector equation AI=E~ is equivalent to the N+1 equations obtained from Ohm's Law and Kirchoff's Law. Then solve the system. Hint: If you are having trouble making the matrix A, try it on paper first for a small N. What do you need to put in the first row of A so that you get the equation I0R0+V=E0 ? What do you need to put in the last row of A to get Kirchoff's law? You are given a number N of branches in the circuit, and two corresponding arrays of shape (Nt) : E which contains the N electromotive forces E0,E1,,EN1, and R which contains the N resistances R0,R1,,RN1. The setup code gives the following variables: Your code snippet should define the following variables: Note: The use of np.linalg. solve() or np.linalg. inv() is not permitted in this question. How else can you solve this system of equations? Hint: sla.lu is still allowed for this question. user_code.py import numpy as np import scipy.linalg as sla \#Step 0: Modify the vector E by appending a zero, so that it looks like the vector E above. \#Step 1: Create the matrix A. A= \#Step 2: Solve the system AI =E. I=