Inverse of a Matrix M as a Product of Elementary Row Operations. Uniqueness?

  • Thread starter Bacle
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Main Question or Discussion Point

Hi, Everyone:

A question about finding the inverse of a matrix M using elementary

row operations (ERO's) E_k (where E_k is either a row-exchange, a scaling

of a row by k, or adding the multiple of one row to another row )

to do row-reduction in reduced-row-echelon

format, to end with the identity, which will have the form :

(E_k*E_(k-1)*....*E_2*E_1)*M=I

My doubt is that these ERO's may be done in different order, and

ERO matrices do not generally commute with each other. Also, the

process of row-reduction may be done in different ways: in some

cases, we may choose, e.g., to swap rows, and in other cases, we

may not. Bottom line is that we may arrive at the identity matrix

I in more than one way, i.e., by applying, in each case, different

sets of ERO's. As a specific example, I am thinking that, in one

choice, we may swap rows if the first row has leading coefficient

larger than 1 to avoid working with fractions, using matrices E_k. In another choice, we

may not swap rows, and work with fractions. In these two cases,

we are using different matrices E_k' to find the inverse of M. How

can we then guarantee that the products:

E_k*E_(k-1)*....*E_2*E_1

E_j' *E'_(j-1)*....*E_2'*E_1'


are equal to each other?

How can we then guarantee that the inverse matrix M that we

get this way-- As the product matrix E_k*E_(k-1)*....*E_2*E_1 is unique--as an inverse

must be?

Thanks in Advance.
 

Answers and Replies

  • #2
AlephZero
Science Advisor
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As you said, the inverse matrix is unique. If B and C satisfy AB = BA = I and AC = CA = I, then
AB = AC
BAB = BAC
IB = IC
B = C

So if you have any numerical method that produces AN inverse matrix, you have found THE (unique) inverse matrix.

Sure, there an many ways to find the inverse with EROs, just like there are many ways to solve linear equations with EROs
 

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