MTH202 Worksheet 6¶

In [1]:
import numpy as np
In [2]:
def RREF(A, showsteps='No'):

    m, n = A.shape
    A = A.astype(float)

    if showsteps=='Yes':
        print(f"Input matrix \nA = \n{A}\n\n")
    
    I = 0
    for j in range(n):
        if showsteps=='Yes':
            print(f"-----CONSIDER COL {j}-----\n")
        for i in range(m):
            if i >= I and A[i,j] != 0:
                if showsteps=='Yes':
                    print(f"CONSIDER ROW {I}\nSWAP rows {i} and {I}:")
                A[[I,i]] = A[[i,I]]
                if showsteps=='Yes':
                    print(f"A = \n{A}")
                A[I,:] /= A[I,j]
                A[I+1:,:] -= A[I+1:,[j]] * A[I,:]
                A[:I,:] -= A[:I,[j]] * A[I,:]   # This line is added from the code in 5.1.
                if showsteps=='Yes':
                    print(f"PIVOT at ({I},{j}): \nA = \n{A}\n\nROW {I} DONE.")
                I += 1
                break
        if showsteps=='Yes':
            print(f"COL {j} DONE.\n")
    
    if showsteps=='Yes':
        print("-----ALL COLs DONE-----\n")
        print(f"RREF(A) = \n{A}")
    
    return A

Problem 1¶

The above block defines a function RREF, which is used to find the RREF of a given matrix.

Instruction
Consider a matrix equation $Ax = b$. Write a function solnum that inputs the matrix A and vector b, then print the number of solutions of $Ax = b$.

Expected behavior

  • When there is a unique solution.
in:  solnum(A,b)
out: Ax = b has a unique solution.
  • When there are multiple solutions.
in:  solnum(A,b)
out: Ax = b has multiple solutions.
  • When there is no solution.
in:  solnum(A,b)
out: Ax = b has no solution.
In [3]:
# Write your function here.

def solnum(A,b):
    """
    The inputs A and b should have matching dimensions.
    The following format should hold.
    A.shape = (m,n) and b.shape = (m,1).
    """
    m, n = A.shape
    A_rref = RREF(A)
    Ab = np.hstack((A,b))
    Ab_rref = RREF(Ab)
    rank_A = 0
    rank_Ab = 0
    for i in range(m):
        if (A_rref[i] != np.zeros((n,))).any():
            rank_A += 1
        if (Ab_rref[i] != np.zeros((n+1,))).any():
            rank_Ab += 1
    print(rank_A)
    print(rank_Ab)
    if rank_A != rank_Ab:
        print("Ax = b has no solution.")
    elif rank_A == rank_Ab and rank_A == n:
        print("Ax = b has a unique solution.")
    else:
        print("Ax = b has multiple solutions.")

Tests¶

In [4]:
A1 = np.array([[1,1,1],
               [0,0,1],
               [0,1,0]])
A2 = np.array([[1,1,1],
               [0,0,0],
               [1,1,1]])
A3 = np.array([[1,1,1],
               [0,0,0],
               [1,0,0]])
b = np.array([[3,0,1]]).T
In [5]:
solnum(A1,b) # unique solution

3
3
Ax = b has a unique solution.
In [6]:
solnum(A2,b) # no solution

1
2
Ax = b has no solution.
In [7]:
solnum(A3,b) # multiple solution

2
2
Ax = b has multiple solutions.