Let: {{begin-eqn}} {{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n       | r = 0       | c =  }} {{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n       | r = 0       | c =  }} {{eqn | o = \vdots }} {{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_{m n} x_n       | r = 0       | c =  }} {{end-eqn}} be the system of equations to be solved. Suppose the elementary row operation of multiplying one row $i$ by a non-zero scalar $\lambda$ is performed. Recall, the $i$th row of the matrix represents the $i$th equation of the system to be solved. Then this is logically equivalent to multiplying the $i$th equation on both sides by the scalar $\lambda$: {{begin-eqn}} {{eqn | l = \alpha_{i 1} x_1 + \alpha_{i 2} x_2 + \ldots + \alpha_{i n} x_n       | r = 0 }} {{eqn | ll= \to       | l = \lambda \alpha_{i 1} x_1 + \lambda \alpha_{i 2} x_2 + \ldots + \lambda \alpha_{i n} x_n       | r = 0       | c = $r_i \to \lambda r_i$ }} {{end-eqn}} which clearly has the same solutions as the original equation. Suppose the elementary row operation of adding a scalar multiple of row $i$ to another row $j$ is performed. Recall that the $i$th and $j$th row of the matrix represent the $i$th and $j$th equation in the system to be solved. {{explain|Woolly. The matrix (by which I presume you mean $\mathbf A$) contains the coefficients and so no part of it "represents" an equation. The act of multiplying $\mathbf x$ by it to obtain $\mathbf b$ represents the equation.}} Thus this is logically equivalent to manipulating the $i$th and $j$th equations as such: {{begin-eqn}} {{eqn | l = \alpha_{i 1} x_1 + \alpha_{i 2} x_2 + \ldots + \alpha_{i n} x_n       | r = 0       | c =  }} {{eqn | l = \alpha_{j 1} x_1 + \alpha_{j 2} x_2 + \ldots + \alpha_{j n} x_n       | r = 0       | c =  }} {{eqn | ll= \to       | l = \alpha_{j 1} x_1 + \alpha_{j 2} x_2 + \ldots + \alpha_{j n} x_n + \lambda \paren {\alpha_{i 1} x_1 + \alpha_{i 2} x_2 + \ldots + \alpha_{i n} x_n}       | r = 0       | c = $r_i \to r_i + \lambda r_j$ }} {{end-eqn}} As both sides of equation $i$ are equal to each other, this operation is simply performing the same act on both sides of equation $j$. This clearly will have no effect on the solution set of the system of equations. Suppose the elementary row operation of interchanging row $i$ and row $j$ is performed. Recall that the $i$th and $j$th row of the matrix represent the $i$th and $j$th equation in the system to be solved. Then, interchanging row $i$ and row $j$ is logically equivalent to switching the $i$th equation and the $j$th equation of the system to be solved. But clearly the system containing the following two equations: {{begin-eqn}} {{eqn | l = \alpha_{i 1} x_1 + \alpha_{i 2} x_2 + \cdots + \alpha_{i n} x_n       | r = 0       | c =  }} {{eqn | l = \alpha_{j 1} x_1 + \alpha_{j 2} x_2 + \cdots + \alpha_{j n} x_n       | r = 0       | c =  }} {{end-eqn}} has the same solution set as a system instead containing the following two equations: {{begin-eqn}} {{eqn | l = \alpha_{j 1} x_1 + \alpha_{j 2} x_2 + \cdots + \alpha_{j n} x_n       | r = 0       | c =  }} {{eqn | l = \alpha_{i 1} x_1 + \alpha_{i 2} x_2 + \cdots + \alpha_{i n} x_n       | r = 0       | c = $r_i \leftrightarrow r_j$ }} {{end-eqn}} Hence the result, by the definition of row equivalence. {{qed}} {{proofread}} Category:Linear Algebra ix7loxpm1evprncf15v9560q41jslzx