# Inverse Matrices

• Dec 30th 2009, 07:01 PM
Inverse Matrices
I'm working through a tutorial on matrices: Pauls Online Notes : Linear Algebra - Finding Inverse Matrices

I'm confused about the following bit:

If $A$ is invertible then there are a set of elementry matrices $E_k...E_2E_1$ such that $E_k...E_2E_1A=I_n$. If we multiply both sides of this by $A^{-1}$:

$E_k...E_2E_1AA^{-1}=I_nA^{-1}\Rightarrow A^{-1}=E_k...E_2E_1I_n$

I'm confused by that statement. I don't see how to get to isolate the inverse to obtain $A^{-1}=E_k...E_2E_1I_n$. On the left side of the arrow, I have $E_k...E_2E_1AA^{-1}=E_k...E_2E_1I=I_nA^{-1}$, but I still don't see how this implies that $A^{-1}=E_k...E_2E_1I_n$ (Headbang)
• Dec 30th 2009, 07:11 PM
tonio
Quote:

Originally Posted by adkinsjr
I'm working through a tutorial on matrices: Pauls Online Notes : Linear Algebra - Finding Inverse Matrices

I'm confused about the following bit:

If $A$ is invertible then there are a set of elementry matrices $E_k...E_2E_1$ such that $E_k...E_2E_1A=I_n$. If we multiply both sides of this by $A^{-1}$:

$E_k...E_2E_1AA^{-1}=I_nA^{-1}\Rightarrow A^{-1}=E_k...E_2E_1I_n$

I'm confused by that statement. I don't see how to get to isolate the inverse to obtain $A^{-1}=E_k...E_2E_1I_n$. On the left side of the arrow, I have $E_k...E_2E_1AA^{-1}=E_k...E_2E_1I=I_nA^{-1}$, but I still don't see how this implies that $A^{-1}=E_k...E_2E_1I_n$ (Headbang)

For any square matrix $K$ of order n, $KI_n=I_nK=K$

Tonio
• Dec 30th 2009, 07:41 PM
Ok, I think I was confused by the notation. I forgot that $I_n$ was just an nxn identity matrix. For some reason I thought it was an elementary matrix with 1 row op.