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Mathematics LibreTexts

23.1: Review the Properties of Invertible Matrices

( \newcommand{\kernel}{\mathrm{null}\,}\)

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%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
from urllib.request import urlretrieve
sym.init_printing(use_unicode=True)
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
            'answercheck.py');
%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
from urllib.request import urlretrieve
sym.init_printing(use_unicode=True)
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
            'answercheck.py');

Let A be an n×n matrix. The following statements are equivalent.

  • The column vectors of A form a basis for Rn
  • |A|0
  • A is invertible.
  • A is row equivalent to In (i.e. it’s reduced row echelon form is In)
  • The system of equations Ax=b has a unique solution.
  • rank(A)=n

Consider the following example. We claim that the following set of vectors form a basis for R3:

B={(2,1,3),(1,6,0),(3,4,10)}

Remember for these two vectors to be a basis they need to obay the following two properties:

  1. They must span R3.
  2. They must be linearly independent.

Using the above statements we can show this is true in multiple ways.

The column vectors of A form a basis for Rn

Do This

Define a numpy matrix A consisting of the vectors B as columns:

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from answercheck import checkanswer

checkanswer.matrix(A,'94827a40ec59c7d767afe6841e1723ce');
from answercheck import checkanswer

checkanswer.matrix(A,'94827a40ec59c7d767afe6841e1723ce');

|A|0

Do This

The first in the above properties tell us that if the vectors in B are truly a basis of R3 then |A|=0. Calculate the determinant of A and store the value in det.

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#Verify that the determinate is in fact zero
if np.isclose(det,0):
    print("Since the Determinate is zero the column vectors do NOT form a Basis")
else:
    print("Since the Determinate is non-zero then the column vectors form a Basis.")
#Verify that the determinate is in fact zero
if np.isclose(det,0):
    print("Since the Determinate is zero the column vectors do NOT form a Basis")
else:
    print("Since the Determinate is non-zero then the column vectors form a Basis.")

A is invertible.

Do This

Since the determinant is non-zero we know that there is an inverse to A. Use python to calculate that inverse and store it in a matrix called A_inv

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from answercheck import checkanswer

checkanswer.matrix(A_inv,'001aaddd4824f42ad9d2ccde21cf9d24');
from answercheck import checkanswer

checkanswer.matrix(A_inv,'001aaddd4824f42ad9d2ccde21cf9d24');

A is row equivalent to In (i.e. it’s reduced row echelon form is In)

Do This

According to the property above the reduced row echelon form of an invertable matrix is the Identiy matrix. Verify using the python sympy library and store the reduced row echelone matrix in a variable called rref if you really need to check it.

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from answercheck import checkanswer

checkanswer.matrix(rref,'cde432847c1c4b6d17cd7bfacc457ed1');
from answercheck import checkanswer

checkanswer.matrix(rref,'cde432847c1c4b6d17cd7bfacc457ed1');

The system of equations Ax=b has a unique solution.

Let us assume some arbitrary vector bRn. According to the above properties it should only have one solution.

Do This

Find the solution to Ax=b for the vector b=(10,200,3). Store the solution in a variable called x

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from answercheck import checkanswer

checkanswer.vector(x,'161cfd16545b1b5fb13e35d2800f13df');
from answercheck import checkanswer

checkanswer.vector(x,'161cfd16545b1b5fb13e35d2800f13df');

rank(A)=n

The final property says that the rank should equal the dimension of Rn. In our example n=3. Find a python function to calculate the rank of A. Store the value in a variable named rank to check your answer.

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#Verify that the determinate is in fact zero
if np.isclose(rank,3):
    print("Rank is 3")
else:
    print("Rank is not 3. Did we do something wrong?")
#Verify that the determinate is in fact zero
if np.isclose(rank,3):
    print("Rank is 3")
else:
    print("Rank is not 3. Did we do something wrong?")
Question (assignment-specific)

Without doing any calculations (i.e. only using the above properties), how many solutions are there to Ax=0? What is(are) the solution(s)?


This page titled 23.1: Review the Properties of Invertible Matrices is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Dirk Colbry via source content that was edited to the style and standards of the LibreTexts platform.

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