# Proof involving invertible matrices

• Feb 21st 2010, 11:36 AM
Proof involving invertible matrices
I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I

3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.

How can I write the above statements so that they flow in a step-by-step proof?
• Feb 21st 2010, 11:40 AM
tonio
Quote:

Originally Posted by madmartigano
I need to prove that if A is an invertible matrix and B is row equivalent to A, then B is also invertible.

I understand the logic of that statement, but I'm not experienced enough to know how to write it into words that prove it.

Here's my thinking so far:

1.) B is row equivalent to A, therefore there exist a sequence of elementary row operations that can be performed on A to produce B:
(En*...*E2*E1)A=B

2.) A is invertible, therefore there exist a sequence of elementary row operations that can be performed on A to produce the identity matrix:
(En*...*E2*E1)A=I

Well, yes, but the first $\displaystyle E_i's$ do not HAVE to be the same as the second ones, do they? So call the elementary matrices representing the row operations in 2. as $\displaystyle F_1,F_2,...,F_m$ (watch the index!), and try again (you're close...!)

Tonio

3.) From 1 & 2, there exist a sequence of elementary row operations that can be performed on B to produce the identity matrix, therefore B is also invertible.

How can I write the above statements so that they flow in a step-by-step proof?

.
• Feb 21st 2010, 12:11 PM
$\displaystyle \begin{array}{l} {E_n} \cdots {E_2}{E_1}A = I\,\,\, \Rightarrow \,\,\,\,A = {E_1}^{ - 1}{E_2}^{ - 1} \cdots {E_n}^{ - 1} \\ \\ {F_m} \cdots {F_2}{F_1}A = B\,\,\, \Rightarrow \,\,\,A = {F_1}^{ - 1}{F_2}^{ - 1} \cdots {F_m}^{ - 1}B \\ \\ {E_1}^{ - 1}{E_2}^{ - 1} \cdots {E_n}^{ - 1} = {F_1}^{ - 1}{F_2}^{ - 1} \cdots {F_m}^{ - 1}B\,\,\, \Rightarrow \,\,\,{E_n} \cdots {E_2}{E_1}B = {F_m} \cdots {F_2}{F_1} \\ \end{array}$
If $\displaystyle E_1E_2\cdot\cdot\cdot E_n A= B$ then $\displaystyle B^{-1}= (E_1E_2\cdot\cdot\cdot E_n A)^{-1}$$\displaystyle = A^{-1}E_n^{-1}\cdot\cdot\cdot\ E_2^{-1}E_1^{-1}$ so B is invertible.