Proof by induction over $\hat n \in \N$, the number of elementary row operations. === Basis for the Induction === Suppose that $\hat n = 0$. Then, the proof is trivial: :$\mathbf C' = \mathbf C = \mathbf A \mathbf B = \mathbf A' \mathbf B$ === Induction Hypothesis === Let $\hat n \in \N$ be fixed. Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf A$. Let $\mathbf C' = \sqbrk {c'}_{m p}$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf C = \mathbf A \mathbf B$. The induction hypothesis is that: :$ \mathbf C' = \mathbf A' \mathbf B$ By definition of matrix product, this is equivalent to: :$\displaystyle \forall i \in \set {1, \ldots, m}, k \in \set {1, \ldots, p}:  c'_{i k} = \sum_{j \mathop = 1}^n a'_{i j} b_{j k}$ === Induction Step === Let $\hat o_1, \ldots, \hat o_{\hat n}, \hat o_{\hat n + 1}$ be a finite sequence of elementary row operations that can be performed on a matrix over $R$ with $m$ rows. Let $\mathbf A'' = \sqbrk {a''}_{m n}$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n + 1}$ on $\mathbf A$. Let $\mathbf C'' = \sqbrk {c''}_{m p}$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n + 1}$ on $\mathbf C$. We must show that $\mathbf C'' = \mathbf A'' \mathbf B$, which by definition of matrix product is equivalent to: :$\displaystyle \forall i \in \set {1, \ldots, m}, k \in \set {1, \ldots, p}: c''_{i k} = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}$ Let $r_{i_0}$ denote the $i_0$'th row of a matrix. It is clear that $\hat o_{\hat n + 1}$ converts $\mathbf A'$ to $\mathbf A''$, and $\hat o_{\hat n + 1}$ converts $\mathbf C'$ to $\mathbf C''$. First, suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \to \alpha r_{i_0}$, where $\alpha \in R$ and $i_0 \in \left\{ {1, \ldots, m}\right\}$. If $i \ne i_0$, then: {{begin-eqn}} {{eqn | ll= (\text i)       | l = c''_{i k}       | r = c'_{i k}       | c = as $\hat o_{\hat n + 1}$ has no effect on $c'_{i k}$ }} {{eqn | r = \sum_{j \mathop = 1}^{n} a'_{i j} b_{j k}       | c = by the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}       | c = as $\hat o_{\hat n + 1}$ has no effect on $a'_{i j}$ }} {{end-eqn}} If $i = i_0$, then: {{begin-eqn}} {{eqn | l = c''_{i k}       | r = \alpha c'_{i_0 k}       | c = by $\hat o_{\hat n + 1}$ }} {{eqn | r = \alpha \sum_{j \mathop = 1}^{n} a'_{i_0 j} b_{j k}       | c = by the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{n} \left({\alpha a'_{i_0 j} }\right) b_{j k} }} {{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}       | c = by $\hat o_{\hat n + 1}$ }} {{end-eqn}} {{qed|lemma}} Suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \to r_{i_0} + \alpha r_{i_1}$, where $\alpha \in R$ and $i_0, i_1 \in \left\{ {1, \ldots, m}\right\}, i_0 \ne i_1$. If $i \ne i_0$, then $\displaystyle c''_{i k} = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}$ as in the equation $(\text i )$. If $i = i_0$, then: {{begin-eqn}} {{eqn | l = c''_{i k}       | r=  c''_{i_0 k} + \alpha c''_{i_1 k}       | c = by $\hat o_{\hat n + 1}$ }} {{eqn | r = \sum_{j \mathop = 1}^n a'_{i_0 j} b_{j k} + \alpha \sum_{j \mathop = 1}^n a'_{i_1 j} b_{j k}       | c = by the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^n \left({a'_{i_0 j} + \alpha a'_{i_1 j} }\right) b_{j k} }} {{eqn | r = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}       | c = by $\hat o_{\hat n + 1}$ }} {{end-eqn}} {{qed|lemma}} Suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \leftrightarrow r_{i_1}$, where $i_0, i_1 \in \set {1, \ldots, m}, i_0 \ne i_1$. If $i \ne i_0$ and $i \ne i_1$, then $\displaystyle c''_{i k} = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}$ as in the equation $(\text i)$. If $i = i_0$, then: {{begin-eqn}} {{eqn | l = c''_{i k}       | r = c'_{i_1 k}       | c = by $\hat o_{\hat n + 1}$ }} {{eqn | r = \sum_{j \mathop = 1}^n a'_{i_1 j} b_{j k}       | c = by the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}       | c = by $\hat o_{\hat n + 1}$ }} {{end-eqn}} If $i = i_1$, then: {{begin-eqn}} {{eqn | l = c''_{i k}       | r = c'_{i_0 k}       | c = by $\hat o_{\hat n + 1}$ }} {{eqn | r = \sum_{j \mathop = 1}^{n} a'_{i_0 j} b_{j k}       | c = by the induction hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}       | c = by $\hat o_{\hat n + 1}$ }} {{end-eqn}} {{qed|lemma}} Then the induction step is proved for all three types of elementary row operations. {{qed}}