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$A$ is a $n\times n$ matrix and there exists a unique matrix $B$ such that $AB=I_n$. Prove that $BA=I_n$ and $B=A^{-1}$.

I have no idea how $BA=I_n$ and $AB=I_n$, one implies another. Please help me to solve.

user1942348
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    What are you allowed to use ? $AB=I$ implies that $\det(A)\det(B)=\det(I)=1$, so both $A$ and $B$ are invertible. Then $AB=I$ gives $B=A^{-1}(AB)=A^{-1}I=A^{-1}$. – Dietrich Burde Nov 22 '15 at 16:28
  • It has something to do with the structure of a (finite-dimensional) matrix ring. The same can't be said for other rings or for infinite-dimensional matrices. That is, for finite-dimensional square matrices $A$ and $B$ of the same size, if $AB=I$, then $A$ is invertible, whence $A^{-1}$ exists, so $B=A^{-1}$, as Dietrich Burde points out. – Batominovski Nov 22 '15 at 16:29
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    Here is a highly upvoted thread for this question http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – littleO Nov 22 '15 at 21:10
  • Multiposted question.(-1). –  Nov 23 '15 at 00:52

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Since $AB=I$ and they are square matrixes, then they both are invertible (various arguments can be done if needed). What we can suppose going wrong is that the left inverse is different from the right inverse. Let us suppose that the right inverse of $A$ is $C$ which means $CA=I$. Then start from \begin{equation} BA=I \end{equation} Simply multiply both term rightly with $A$ and get $$ABA=A$$ than left-multiply with $C$ and obtain \begin{equation} CABA=CA \end{equation} But since $CA=I$ last equation lead to \begin{equation} BA=I \end{equation} As corollary we obtain that $A^{-1}=C=B$.

Dac0
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