The Inverse Matrix is Unique

Problems and solutions in Linear Algebra

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 $A$.


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