Inverse Matrices

**adkinsjr** · December 30th 2009, 07:01 PM

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 $A^{-1}=E_k...E_2E_1I_n$

**tonio** · December 30th 2009, 07:11 PM

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 $A^{-1}=E_k...E_2E_1I_n$

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

Tonio

**adkinsjr** · December 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.

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