$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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proof
If $A$ is $n\times n$ and has a left inverse $B$ and a right inverse $C$, then $B=C$. Furthermore, the inverse $A^{-1}$ is unique.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$$\begin{align*} B&= BI\\ &=B(AC) \\ &=(BA)C \\ &=IC\\ &=C \end{align*}$$
If $A$ is $n\times n$ and has a left inverse $B$ and a right inverse $C$, then $B=C$. Furthermore, the inverse $A^{-1}$ is unique.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$$\begin{align*} B&= BI\\ &=B(AC) \\ &=(BA)C \\ &=IC\\ &=C \end{align*}$$
FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof
If $A$ is $n\times n$ and has a left inverse $B$ and a right inverse $C$, then $B=C$. Furthermore, the inverse $A^{-1}$ is unique.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$$\begin{align*} B&= BI\\ &=B(AC) \\ &=(BA)C \\ &=IC\\ &=C \end{align*}$$
If $A$ is $n\times n$ and has a left inverse $B$ and a right inverse $C$, then $B=C$. Furthermore, the inverse $A^{-1}$ is unique.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

$$\begin{align*} B&= BI\\ &=B(AC) \\ &=(BA)C \\ &=IC\\ &=C \end{align*}$$
FullPage
result
concepts
hypothesis
implications
proof