# The Inverse Matrix is Unique

## Problem 251

Let $A$ be an $n\times n$ invertible matrix. Prove that the inverse matrix of $A$ is uniques.

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## Hint.

That the inverse matrix of $A$ is unique means that there is only one inverse matrix of $A$.
(That’s why we say “the” inverse matrix of $A$ and denote it by $A^{-1}$.)
So to prove the uniqueness, suppose that you have two inverse matrices $B$ and $C$ and show that in fact $B=C$.

Recall that $B$ is the inverse matrix if it satisfies
$AB=BA=I,$ where $I$ is the identity matrix.

## Proof.

Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. Then they satisfy
$AB=BA=I \tag{*}$ and
$AC=CA=I \tag{**}.$

To show that the uniqueness of the inverse matrix, we show that $B=C$ as follows. Let $I$ be the $n\times n$ identity matrix.
We have
\begin{align*}
B&=BI\\
&=B(AC) && \text{ by (**)}\\
&=(BA)C &&\text{ by the associativity}\\
&=IC && \text{ by (*)}\\
&=C.
\end{align*}
Thus, we must have $B=C$, and there is only one inverse matrix of $A$.

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that...