One method to put a matrix A into reduced row echelon form (RRE):
NOTE: The leading entry of a row of a matrix is the left-most non- zero entry
of that row.
1. Be clever. Then, put all rows consisting entirely of zeros at the bottom
of the matrix.
2. Choose a simple row with a leading entry in the first column. Switch this row
with the first
row. Use this row and elementary row ope rations to get all entries in the first
column below
the first row equal to zero. (I say clean down the first column).
3. Now, forget about this first row. Choose a simple row with a leading entry in
the second
column. Switch this row with the second row. Use this row and elementary row
operations to
make all entries be low its leading entry equal to zero (clean down the second
column).
4. Keep cleaning down successive columns.
The matrix is now in row echelon form (but not necessary reduced row echelon
form). The
leading entries are in the pivot positions for A.
5. Clean up. That is, use the lowest non-zero row and elementary row operators
to make all
entries in the column above its leading entry zero. Make that leading entry
equal to one using
a row operation.
6. Now, forget about this lowest row and use the row above it to clean up,
making all entries in
the column above its leading entry zero. Then, make its leading entry equal to
one using a
row operation.
7. Continue cleaning up. Then, make sure all leading entries are equal to one
and the matrix is
in RRE.
To solve m × n linear algebraic equations Ax = b:
1. Form the augmented matrix [Ab] and reduce to RRE.
2a. If in this process you get a leading entry in the last column (the matrix
has a row: (0, . . . , 0, b)
where b ≠ 0) then the system is inconsistent . (That is,
there are no solutions to the equations
because one of the equations is 0 = b.)
2b. If not, continue to reduce to RRE, convert to equations and solve for the
leading variables (or
basic variables) ( variables corresponding to leading entries of the reduced
matrix) in terms of
the free variables (non-leading variables) and constants. Note that the leading
(basic) variables
correspond to pivot columns of A.
To solve the m × n homogeneous linear algebraic equations Ax = 0:
1. Reduce the augmented matrix [A0] to RRE. (You can reduce A as long as you
convert back
to homogeneous equations.)
2. Solve for the basic (leading) variables in terms of free variables (if there
are free variables).
3. If there are free variables: successively set each free variable equal to one
(or an ‘easy’ non-
zero number ) and set the other free variables equal to zero. Then, write your
solutions as
vectors. This gives a list of linearly independent solution vectors u1, . . .
,um(one for each
free variable). Every solution to Ax = 0 is a linear combination of these
solution vectors;
therefore, the solution set to Ax = 0 is span{u1, . . . ,um}.
