23.1: Review the Properties of Invertible Matrices
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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:
- They must span R3.
- 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
Define a numpy matrix A
consisting of the vectors B as columns:
|A|≠0
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
.
A is invertible.
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
A is row equivalent to In (i.e. it’s reduced row echelon form is In)
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.
The system of equations Ax=b has a unique solution.
Let us assume some arbitrary vector b∈Rn. According to the above properties it should only have one solution.
Find the solution to Ax=b for the vector b=(−10,200,3). Store the solution in a variable called x
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.
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)?