The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$ === Edge Cases === ==== $n = 1$ ==== Consider the case where $n = 1$. Then: {{begin-eqn}} {{eqn | l = \mathbf {A B}       | r = a_{11} b_{11} }} {{eqn | l = \mathbf {B A}       | r = b_{11} a_{11}       | c =  }} {{end-eqn}} and it follows that (conventional) matrix multiplication over $\map {\MM_R} 1$ is commutative {{iff}} $R$ is a commutative ring. ==== $R$ not a Ring with Unity ==== Consider the case where $R$ is not a ring with unity, and is a general ring. Let $R$ be the trivial ring. From Matrix Multiplication on Square Matrices over Trivial Ring is Commutative: :$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$ Hence the result does not follow for all rings. It is not established at this point on exactly which rings (conventional) matrix multiplication $\map {\MM_R} n$ commutes. However, the existence of just one such ring (the trivial ring) warns us that we cannot apply the main result to ''all'' rings. ==== Matrices are not Square ==== We note that $\mathbf A \mathbf B$ is defined when: :$\mathbf A = \sqbrk a_{m n}$ is an $m \times n$ matrix :$\mathbf B = \sqbrk b_{n p}$ is an $n \times p$ matrix. Hence for both $\mathbf A \mathbf B$ and $\mathbf B \mathbf A$ to be defined, it is necessary that: :$\mathbf A = \sqbrk a_{m n}$ is an $m \times n$ matrix :$\mathbf B = \sqbrk b_{n p}$ is an $n \times m$ matrix for some $m, n \in \Z_{>0}$. But in this situation: :$\mathbf A \mathbf B$ is an $m \times m$ matrix while: :$\mathbf B \mathbf A$ is an $n \times n$ matrix and so if $\mathbf A$ and $\mathbf B$ are not square matrices, they cannot commute. === Basis for the Induction === From Matrix Multiplication is not Commutative/Order 2 Square Matrices, it is seen that there exist $2 \times 2$ matrices that don't commute under matrix multiplication, thus proving the result for $n = 2$. $\map P 2$ is the case: :$\exists \mathbf A, \mathbf B \in \map {\MM_R} 2: \mathbf {A B} \ne \mathbf {B A}$ This is demonstrated in Matrix Multiplication is not Commutative: Order $2$ Square Matrices. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\exists \mathbf A, \mathbf B \in \map {\MM_R} k: \mathbf {A B} \ne \mathbf {B A}$ from which it is to be shown that: :$\exists \mathbf A, \mathbf B \in \map {\MM_R} {k + 1}: \mathbf {A B} \ne \mathbf {B A}$ === Induction Step === This is the induction step: From the induction hypothesis, it is assumed that there exist $2$ order $k$ square matrices $\mathbf A$ and $\mathbf B$ such that $\mathbf {A B} \ne \mathbf {B A}$. For an order $n$ square matrix  $\mathbf D$, let $\mathbf {D'}$ be the square matrix of order $n + 1$ defined as: :$d'_{i j} = \begin {cases} d_{i j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$ Thus $\mathbf D'$ is just $\mathbf D$ with a zero row and zero column added at the ends. We have that $\mathbf D$ is a submatrix of $\mathbf D'$. Now: :$\paren {a' b'}_{i j} = \begin{cases} \displaystyle \sum_{r \mathop = 1}^{n + 1} \mathbf a'_{i r} b'_{r j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$ But: {{begin-eqn}} {{eqn | l = \sum_{r \mathop = 1}^{n + 1} a'_{i r} b'_{r j}       | r = a'_{i \paren {n + 1} } b'_{\paren {n + 1} i} + \sum_{r \mathop = 1}^n a'_{i r} b'_{r j}       | c =  }} {{eqn | r = \sum_{r \mathop = 1}^n a_{i r} b_{r j}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \mathbf A' \mathbf B' \paren {n + 1, n + 1}       | r = \paren {\mathbf {A B} }' \paren {n + 1, n + 1}       | c =  }} {{eqn | r = \mathbf {A B}       | c =  }} {{eqn | o = \ne       | r = \mathbf {B A}       | c =  }} {{eqn | r = \paren {\mathbf {B A} }' \paren {n + 1, n + 1}       | c =  }} {{eqn | r = \mathbf B' \mathbf A' \paren {n + 1; n + 1}       | c =  }} {{end-eqn}} Thus it is seen that: :$\exists \mathbf A', \mathbf B' \in \MM_{n + 1 \times n + 1}: \mathbf A' \mathbf B' \ne \mathbf B' \mathbf A'$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$ and by definition (conventional) matrix multiplication over $\map {\MM_R} n$ is not commutative. {{qed}}