## Instructions

**Objective**

Write a python assignment to compute matrix computations.

## Requirements and Specifications

**Source Code**

```
import numpy as np
import warnings
def swapRows(A, i, j):
"""
interchange two rows of A
operates on A in place
"""
tmp = A[i].copy()
A[i] = A[j]
A[j] = tmp
def relError(a, b):
"""
compute the relative error of a and b
"""
with warnings.catch_warnings():
warnings.simplefilter("error")
try:
return np.abs(a-b)/np.max(np.abs(np.array([a, b])))
except:
return 0.0
def rowReduce(A, i, j, pivot):
"""
reduce row j using row i with pivot pivot, in matrix A
operates on A in place
"""
factor = A[j][pivot] / A[i][pivot]
for k in range(len(A[j])):
# we allow an accumulation of error 100 times larger than a single computation
# this is crude but works for computations without a large dynamic range
if relError(A[j][k], factor * A[i][k]) < 100 * np.finfo('float').resolution:
A[j][k] = 0.0
else:
A[j][k] = A[j][k] - factor * A[i][k]
# stage 1 (forward elimination)
def forwardElimination(B):
"""
Return the row echelon form of B
"""
A = B.copy().astype(float)
m, n = np.shape(A)
for i in range(m-1):
# Let lefmostNonZeroCol be the position of the leftmost nonzero value
# in row i or any row below it
leftmostNonZeroRow = m
leftmostNonZeroCol = n
## for each row below row i (including row i)
for h in range(i,m):
## search, starting from the left, for the first nonzero
for k in range(i,n):
if (A[h][k] != 0.0) and (k < leftmostNonZeroCol):
leftmostNonZeroRow = h
leftmostNonZeroCol = k
break
# if there is no such position, stop
if leftmostNonZeroRow == m:
break
# If the leftmostNonZeroCol in row i is zero, swap this row
# with a row below it
# to make that position nonzero. This creates a pivot in that position.
if (leftmostNonZeroRow > i):
swapRows(A, leftmostNonZeroRow, i)
# Use row reduction operations to create zeros in all positions
# below the pivot.
for h in range(i+1,m):
rowReduce(A, i, h, leftmostNonZeroCol)
return A
####################
# If any operation creates a row that is all zeros except the last element,
# the system is inconsistent; stop.
def inconsistentSystem(A):
"""
B is assumed to be in echelon form; return True if it represents
an inconsistent system, and False otherwise
"""
m, n = np.shape(A)
for i in range(m):
for j in range(n):
if (A[i][j] != 0):
if (j == n-1):
return True
else:
break
return False
def backsubstitution(B):
"""
return the reduced row echelon form matrix of B
"""
A = B.copy().astype(float)
m, n = np.shape(A)
for i in range(m):
# If row i is all zeros, or if i exceeds the number of rows in A, stop.
for j in range(n):
if A[i][j] != 0.0:
break
if j == n-1:
return A
pivot = j
# If row i has a nonzero pivot value, divide row i by its pivot value.
# This creates a 1 in the pivot position.
A[i] = A[i] / A[i][pivot]
for j in range(i+1, m):
rowReduce(A, i, j, pivot)
return A
#####################
print("N (8 pts)")
B = backsubstitution(np.array([[-25, -29, -27, 1, 0, 0], [546, 180, 537, 0, 1, 0], [154, 50, 149, 0, 0, 1]]))
swapRows(B, 0, 2)
B = backsubstitution(B)
swapRows(B, 0, 2)
A2 = B[0:3, 4]
A3 = B[0:3, 5]
print("Second column of inversed matrix:", A2)
print("Third column of inversed matrix:", A3)
print("\n\n")
print("O (5 pts)")
print("Let A is a matrix of rearrangement, mentioned in the assignment")
A = [[0.9, 0.01, 0.09], [0.01, 0.9, 0.01], [0.09, 0.09, 0.9]]
print("A =",A)
print("Let vector X represents initial distribution of bike fleet among 3 locations (i.e. [1000, 1000, 500])")
print("Then A*X gives us distribution after first dat, A^2*X gives distribution after 2 days.... A^k*2 - distribution after k days")
print("Let's calculate A^100")
deg = np.eye(3)
for _ in range(100):
deg = np.matmul(deg, A)
print("A^100 =")
print(deg)
print("We see that each row contains approximately equal numbers. It is the property of stochastic matrix (A is stochastic):")
print("A^k converges to matrix with constant value in each row")
print("When we calculate product R = A^100*X, we obtain the following matrix R (for any valid initial vector X):")
fleet = 2500
R = [deg[0][0] * 2500, deg[1][0] * 2500, deg[2][0] * 2500]
print("R =", R)
print("So, the expected number of bikes in downtown is approx", int(R[0]), ", expected number of bikes in Back Bay is approx", int(R[1]),
"expected number of bikes in Other locations is approx", int(R[2]))
Contact Details
Address:
5209 Cedar Glenn
```

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