 Solution

For $i\ne j$, put $x_{ij}(a)=I+a{E}_{ij}$. When $a\ne 0$ this is a transvection: it fixes the hyperplane $v_j=0$ pointwise and has rank-one difference from the identity. Its inverse is $x_{ij}(-a)$ and its determinant is one.

Left multiplication performs a row addition. For an invertible matrix, bring a nonzero entry into a pivot position by a signed interchange, then clear its column by row additions. A signed interchange on two coordinates is itself a product of row additions, becauseRepeating the pivot construction and then clearing above the pivots reduces any determinant-one matrix to a diagonal matrix of determinant one. FinallyA determinant-one diagonal matrix is a product of such two-coordinate matrices, using the last coordinate to compensate each of the first $n-1$ diagonal entries. Each displayed matrix is a product of elementary transvection matrices. Hence reversing the elimination provesThis argument works over every field, including characteristic two.