A One Side Inverse Matrix is the Inverse Matrix: If $AB=I$, then $BA=I$ Problem 548

An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that

1. $AB=I$, and
2. $BA=I$,

where $I$ is the $n\times n$ identity matrix.

If such a matrix $B$ exists, then it is known to be unique and called the inverse matrix of $A$, denoted by $A^{-1}$.

In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.

Prove that $BA=I$, and hence $A^{-1}=B$. Add to solve later

Contents

Proof.

Since $AB=I$, we have
\begin{align*}
\det(A)\det(B)=\det(AB)=\det(I)=1.
\end{align*}
This implies that the determinants $\det(A)$ and $\det(B)$ are not zero.
Hence $A, B$ are invertible matrices: $A^{-1}, B^{-1}$ exist.

Now we compute
\begin{align*}
I&=BB^{-1}=BIB^{-1}\\
&=B(AB)B^{-1} &&\text{since $AB=I$}\\
&=BAI=BA.
\end{align*}
Hence we obtain $BA=I$.
Since $AB=I$ and $BA=I$, we conclude that $B=A^{-1}$. Add to solve later

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