Page 60 - A Course in Linear Algebra with Applications

44 Chapter Two: Systems of Linear Equations

Hence the general solution given by back substitution is x\ =
—2 — c — 3d, X2 = d, £3 = 1 — c/3, £4 = c, where c and d are
arbitrary scalars.
The matrix formulation enables us to put our conclusions
about linear systems in a succinct form.

Theorem 2.2.1
Let AX = B be a linear system of equations in n unknowns
with augmented matrix M = [A \ B].
(i) The linear system is consistent if and only if the matri-
ces A and M have the same numbers of pivots in row echelon
form.
(ii) If the linear system is consistent and r denotes the
number of pivots of A in row echelon form, then the n — r
unknowns that correspond to columns of A not containing a
pivot can be given arbitrary values. Thus the system has a
unique solution if and only if r = n.

Proof
For the linear system to be consistent, the row echelon form
of M must have only zero entries in the last column below the
final pivot; but this is just the condition for A and M to have
the same numbers of pivots.
Finally, if the linear system is consistent, the unknowns
corresponding to columns that do not contain pivots may be
given arbitrary values and the remaining unknowns found by
back substitution.

Reduced row echelon form
A matrix is said to be in reduced row echelon form if it is
in row echelon form and if in each column containing a pivot
all entries other than the pivot itself are zero.