If ~m # ~n matrix A is row equivalent to a reduced echelon matrix D then A &tilde. D, so rank A = rank D = &rho. say. _ Now &exist. &rho. linearly independent rows in D, but the rows of D which are not all zero are linearly independent.
So rank A = number of non-zero rows of D.
Furthermore we know that A &tilde. C, where
C _ = _ matrix{I_{&rho.},0/0,0}
But D is column eqivalent to C, i.e. &exist. elementary column vectors K_1 ... K_{~h} such that C = D K_1 ... K_{~h}.
Now D = R_{~k} ... R_1 A, _ so _ C = R_{~k} ... R_1 A K_1 ... K_{~h} _ but _ C = M A N, where M and N are regular. All the elementary matrices are regular so we can put M = R_{~k} ... R_1, and N = K_1 ... K_{~h}
Example:
In a previous example we showed that
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The ~n # ~n regular matrix A is row equivalent to a ~n # ~n reduced echelon matrix C, but as rank A = rank C = &rho. (see above) then C must have ~n non-zero rows, i.e. C = I_{~n}. _ So _ I_{~n} = R_{~k} ... R_1 A, _ in which case _ R_{~k} ... R_1 = A^{&minus.1}. _ This is a useful way of calculating inverses.
Example: