#Question 1 #x = (ATA)**-1ATb #A = [[1,0],[0,1],[1,1]], AT = [[1,0,1],[0,1,1]], b = [[-1],[1],[0]] #ATA = [[2,1],[1,2]] #(ATA)**-1 = (1/(2*2-1*1))*[[2,-1],[-1,2]] = [[2/3,-1/3],[-1/3,2/3]] #(ATA)**-1AT = [[2/3,-1/3,1/3],[-1/3,2/3,1/3]] #(ATA)**-1ATb = [[-1],[1]]

# Code for Question 1 import numpy as np A = np.array([[1,0],[0,1],[1,1]]) b = np.array([-1,1,0]) AT = A.transpose() Z = np.dot(AT, A) InverseZ = np.linalg.inv(Z) P = np.dot(InverseZ, AT) X = np.dot(P,b) X

#Question 2 #It is not a linear transformation because it does not satisfy the rule of # T(0) = 0. In this case, T(0,0,0)= (1,0,0).

#3Question 3 #a. T(e1)=T([1,0])=[2,0], T(e2) = T([0,1]) = [0,3], # so the martix = [[2,0],[0,3]] #b. Yes, an T2, [[1/2,0],[0,1/3]], is identified as the inverse of the T, # so T is invertible. #c. Ker(T) = span{(0,0)} #d. Range(T) = R2