27. The general solution of a homogeneous system
Denote by r the Rank of the coefficient matrix A of the system of equations Ax=0. According to Theorem 12, there exist among the rows of A exactly r linearly independent rows. In general, it will not be affected by the assumption that the first r rows of A are linearly independent, because, if that should not be so at the start, it is sufficient to reorder the sequence of the rows, which is obviously without consequence for the solutions. Moreover, by Theorem 12, all rows of A depend linearly on the first r rows, whence all equations of the system Ax = 0 are linear combinations of the first r equations.As a consequence, every solution of the first r equations is also a solution of the remaining equations, so that we need only determine the general solution of the first r equations
a11x1
+ ··· + a1nxn
= 0,
····················································
(27.1)
ar1x1 + ··· +
arnxn =
0.
This matrix of coefficients still has the rank r, that is, there exists at least one minor determinant of order r which differs from zero. In order to simplify the notation, we will assume that the minor determinant

This does not restrict the generality, but at the most renumbers the unknowns.
We write now the system (27.1) in the form
a11x1
+ ··· + a1rxr
= -a1,r+1xr+1
- ··· - a1nxn,
····················································
(27.3)
ar1x1 + ··· +
arrxr =
-ar,r+1xr+1
- ··· - arnxn.
If you substitute here arbitrary values for the last n - r unknowns xr+1 ··· xn, you can always, as a result of (27.2), determine by Cramer's Rule x1, ···, xr uniquely, so that Equations (27.3) are fulfilled.
To start with, let
xr+1 = 1, xr+2 = ··· = xn = 0
and find from (27.3) the corresponding values of x1, ···, xr. Let them be
x1 = x1(1), ··· , xr = xr(1).
and denote this solution of (27.1) by x(1). Now let
xr+1 = 0, xr+2 = 1, xr+r = ··· = xn = 0
and find the first r unknowns
x1 = x1(2), ··· , xr = xr(2)
so that you obtain a solution x(2) of (27.3). You can continue in this manner: You set one of the n - r last unknowns equal to 1, the others equal to 0 and determine then the first r unknowns in such a manner that Equations (27.3) are fulfilled. In this manner, you obtain n - r solutions

of the system (27.1). A linear combination
z = l1x(1)+ l2x(2) + ··· + ln-rx(n-r)
has the form

with
zi = l1xi (1)+ l2xi (2) + ··· + ln-rxi (n-r).
To start with, this shows that the solutions (27.4) are linearly independent, because you can only obtain z = 0 with
l1 = l2 = ··· = ln-r = 0.
Moreover, you can obtain every solution of (27.1) as a linear combination of the solutions (27.4), because, as we have seen above, every solution of (27.1) is already uniquely determined by selection of the last n - r unknowns; obviously, you can form by a suitable choice of the l1, ··· , ln a solution z in which the last unknowns have arbitrary values.
If in a given system of equations the minor determinant (27.2) is not zero, but any other one and you do not want to renumber the unknowns, you must then shift during the transition from (27.1) to (27.3) these n - r unknowns to the right hand side, in order to obtain on the left hand side of (27.3) a matrix with a non-zero determinant.
These results are summarized in
If the matrix A of the homogeneous system of equations Ax = 0 with n unknowns has Rank r, there exist n - r linearly independent solutions.
The following observation proves frequently to be useful: The computation of the solutions (27.4) involves only the four basic operations with the coefficients of the equations aik, whence you can always obtain n - r linearly independent solutions, the elements of which are contained in the smallest body of numbers comprising the coefficients aik. Thus, for example, if all the aik are real (rational) numbers, then there exist already in the field of real (rational) numbers n - r linearly independent solutions.
Frequently, you have n - 1 linearly independent equations with n unknowns
b11x1
+ ··· + b1nxn
= 0,
····················································
(27.5)
bn-1x1 + ···
+ bn-1xn =
0
with a matrix B = (bik) of Rank n - 1. Then, by Theorem 14, all solutions are multiples of a single one. Such a solution is readily written down. Let Bi denote the determinant of that matrix, which arises from B by omission of the i-th column. Then the required solution is
x1 = B1, x2 = -B2, ··· , xk = (-1)k+1Bk, ··· , xn = (-1)n+1Bn. (27.6)
Indeed, for i = 1, ··· , n - 1,

because this determinant contains two equal rows. If you develop it with respect to the first row, you obtain
b11B1 +b12(-B2) + ··· + b1k (-1)k+1Bk+ ··· + bin·(-1)n+1Bn = 0 (i = 1,···,n-1),
whence (27.6) is a solution of the system (27.5).
Assume now that the coefficients of Equations (27.1) are real. Denote the rows of the coefficient matrix by z'1, ··· , zr'. Then the system 27.1) can be rewritten in the form
z'1x = 0, ··· , z'r x = 0. (27.7)
These equations state that the solution vector x is orthogonal to the r vectors z1, ··· , zr *. Then z is also orthogonal to all vectors of the r-dimensional vector space Z = z1, ··· , zr, whence follows from (27.7) for arbitrary m1, ··· , m1r
(m1z1 + ··· + mrzr)'x = m1z'1x + ··· + mrz'rx) = 0.
* Solely in order to be in agreement with the concept of orthogonality, we have confined our consideration to real numbers. The reader will readily work out corresponding reflections in the complex regime.
The set of all solutions x form an (n - r)-dimensional vector space L. In this interpretation of (27.1), it will be appropriate to describe L as the vector space, orthogonal to Z. Theoerm 14 can also be given the form:
If Z is any partial space of the n-dimensional vector space V and L is the partial space, orthogonal with respect to Z, then the sum of the dimensions of Z and L equals n.
Conversely, if you start from L and determine the partial space of V, orthogonal to L, you obtain Z, for, on the one hand, Z is orthogonal to L, on the other hand, Z has already the right dimension, which in combination with L yields the sum n.