Inverse: which is the order of multiplying the matrices?

[Minz]

Hi Can't find an explanation anywhere..maybe you'll help?
Given 2 martrices prove A has inverse of A-1. Easy enough,multiply out and get I. But then question what is the inverse of A-1, Explain.

    1 0 59           1 0 -59
A = 0 1 89     A-1 = 0 1 -89
    0 0  1           0 0   1

Is it that AA-1= A-1A=I??

 

[soroban]

Re: Inverse

Hello, Minz!

Given two martrices, prove \(\displaystyle A\) has inverse of \(\displaystyle A^{-1}\)
Easy enough, multiply out and get \(\displaystyle I.\)

But then: What is the inverse of \(\displaystyle A^{-1}\)? . Explain.

\(\displaystyle A \:=\:\begin{pmatrix} 1 & 0 & 59\\0&1&89\\0&0&1\end{pmatrix}\;\;\;A^{-1}\:=\:\begin{pmatrix}1 & 0 & -59\\0 & 1&-89 \\ 0 & 0 & 1\end{pmatrix}\)

Is it that \(\displaystyle A\cdot A^{-1} \:= \:A^{-1}\cdot A\:=\:I\)? . Yes!

Definition of inverse:
. . If \(\displaystyle A\cdot B \,=\,I\), then \(\displaystyle B\) is the inverse of \(\displaystyle A.\)
. . (The second matrix is the inverse of the first matrix.)


We find that: \(\displaystyle \,A\cdot A^{-1} \,=\,I\)
. . Hence, \(\displaystyle A^{-1}\) is the inverse of \(\displaystyle A.\)

We also find that: \(\displaystyle A^{-1}\cdot A \,=\,I\)
. . Hence, \(\displaystyle A\) is the inverse of \(\displaystyle A^{-1}.\)

In other words: \(\displaystyle \:\left(A^{-1}\right)^{-1} \,=\,A\)

 

[Minz]

Thanks

Thanks soroban, that cleared it up.