Reduced Row Echelon Form

The rank of a matrix A is the number of nonzero rows in the reduced row echelon form of A. 

This is a system of three equations in three unknowns (Xj, X2, and X3), which can be solved by standard methods. Let us perform row operations on the above system to make the solution easier to interpret. Hence, performing elementary row operations and reducing the system to only elements on the main diagonal (also known as reduced row echelon form) gives... 

A is a 3 X 5 matrix and thus the null space of A will be a two-dimensional subspace in c -Cb-Cc-Cd-Ce space (the size of matrix N must he nx(n- d), or 5x2). To compute the null space of this matrix, we can reduce A to reduced row echelon form by performing elementary row operations on A, and determine all of the vectors in the null space (similar in procedure to that shown in Example 3). Hence reducing A to the equivalent matrix gives ... 

An augmented molecular matrix can be transformed to a Reduced Row Echelon Form or RREF. This method is essential to all matrix transformations in this chapter. The idea behind the RREF is that we work from the first colunrn all the way to the rightmost one. For each column we determine whether it is possible to eliminate it by finding a nonzero entry, or pivot, in a row that has not been considered before. If not, we skip to the next column. If a pivot is found, we use it to eliminate all other entries in that row. We also move the pivot row up as far as possible. We cannot tell in advance where all the pivots will be found we must find them one by one since the elimination procedure can change zero entries into nonzero ones and vice versa. In general, we also do not know in advance how many pivots will be found. However, in the special case of a matrix augmented with a unit matrix, we do know that their number will be equal to the number of rows. 

The first (preceding) matrix is reduced to echelon form (zeros in the first and the second rows of column one) by... 

In this case, A can be transformed by elementary row operations (multiply the second row by 1/2 and subtract the first row from the result) to the unit-matrix or reduced I0W-echelon form ... 

Of course, first we need to find an initial minimal DD pair. Following the null space approach [23, 24], we compute a basis of the kernel of the stoichiometric matrix S. More specifically, we compute a column-reduced echelon form of the basis and (after a permutation of rows) obtain... 

The matrix U has rank r, so there will be r basic variables and (n - r) free variables in the solution for h. In fact, we may further reduce the system in Equation 5.72 into row-reduced echelon form as follows ... 

Note that permutations of the columns of U may be necessary to obtain the row-reduced echelon form shown in Equation 5.73. Of course, in order to maintain consistency in the equations, column permutations in U must be accompanied by corresponding row permutations in h =. 

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