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First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$ $\mathbf{y}$, the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Indeed, if $\mathbf c$ is a solution then \begin{gather*} \mathbf y = B\mathbf c\\ A\mathbf y = (AB)\mathbf c = \mathbf c \end{gather*}

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$

First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Consider the map $\mathbf x\mapsto B\mathbf x$. If $\mathbf x$ and $\mathbf x'$ are $n\times 1$ matrices, then $B\mathbf x = B\mathbf x'$ implies \begin{align*} AB \mathbf x &= AB \mathbf x' \\ E \mathbf x &= E \mathbf x' \\ \mathbf x &= \mathbf x', \end{align*} that is, left multiplication by $B$ is an injective map. By the rank-nullity theorem, this map is also onto, then for each matrix $n\times 1$ $\mathbf{y}$, the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$

First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$, $\mathbf{y}$ the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Indeed, if $\mathbf c$ is solution then \begin{gather*} \mathbf y = B\mathbf c\\ A\mathbf y = (AB)\mathbf c = \mathbf c \end{gather*}

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$

First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$ $\mathbf{y}$, the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Indeed, if $\mathbf c$ is a solution then \begin{gather*} \mathbf y = B\mathbf c\\ A\mathbf y = (AB)\mathbf c = \mathbf c \end{gather*}

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$

In the light of the uniqueness of the left inverse of a matrix it's possible to give an elementary proof.

First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th comlumn of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$, $\mathbf{y}$ the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, this says that $$C = E.$$

First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$, $\mathbf{y}$ the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Indeed, if $\mathbf c$ is solution then \begin{gather*} \mathbf y = B\mathbf c\\ A\mathbf y = (AB)\mathbf c = \mathbf c \end{gather*}

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$