Is there some special method to find the inverse of an elementary matrix?
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OpenStudy (samiam):
haha u r here without a notification
OpenStudy (anonymous):
Gaussian Row-Reduction is useful. For a two by two matrix,
\[A =\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\]
The inverse matrix is
\[\frac{1}{ad-bc} \left[\begin{matrix}d & -b\\ -c &a\end{matrix}\right]\]
OpenStudy (samiam):
i think they are looking for something using elementary row operations
OpenStudy (samiam):
I mean elentary matrices
OpenStudy (anonymous):
I don't know what you mean by an elementary matrix.
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OpenStudy (samiam):
LOL neither do I ok i will read some more and try to figure oout
OpenStudy (anonymous):
Oh oh oh nevermind, sorry, my fault. You only need one row operation to change an elementary matrix into the identity.
OpenStudy (samiam):
ok
OpenStudy (anonymous):
An example might be
\[\left[\begin{matrix}1 & 0 \\0 & 2\end{matrix}\right]\]
OpenStudy (samiam):
so the second row must be multiplied by a 1/2
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ya but like I am trying to find the inverse of this
OpenStudy (samiam):
using products of elementary matrices
OpenStudy (anonymous):
That's easy to do, just use the usual technique of finding inverse.
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OpenStudy (anonymous):
How about
\[\left[\begin{matrix}1 & 0 \\0 & 1/2\end{matrix}\right]\]
?
OpenStudy (anonymous):
If A is an elementary square matrix.
OpenStudy (samiam):
y did we just do that?
OpenStudy (samiam):
ohhhh cuz we are going backwards?
OpenStudy (anonymous):
Now write \( A= I_n A \)
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OpenStudy (samiam):
Not getting this stuff at all
OpenStudy (anonymous):
Then do sequence of row operation on the L.H.S and the prefactor I_n till we obtain \( I_n=BA \)
This step will be easy since A is an elementary matrix.
OpenStudy (anonymous):
Now we can write \( A^{-1}=B \)
OpenStudy (samiam):
ok so show sth an example plz
OpenStudy (anonymous):
I am too tired for a Latex example now :(
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