General Solutions of Systems in Vector Form MA 2071

We are looking for solutions to the system Ax = b, in column vector form in what follows. We wish to organize the vectors making up the solution into two types, what are called homogeneous and particular.

First, assume we have begun with the augmented matrix (A | b) and use Gauss-Jordan elimination, either by hand or with software, to get it down to "Final Form".

Suppose for sake of discussion the Final Form is equivalent, in equation form, to

x + 3z - 2v = 6

y -9z -13v = 12

w + 7v = 9 (changed from 19 in class)

From last week, the rank is 3, there are 5 variables, so two of them will be arbitrary and will be z and v as they were never used as pivots for elimination purposes.

The general solution, in scalar form, is then

x = -3z + 2v + 6

y = 9z + 13v + 12

z arb

w = - 7v + 9

v arb

This was covered last week. From here, we put all 5 variables into a column vector, in order, x,y,z,w,v:

Next we break it up into 3 vectors, the one with all z's, the one with all v's and the one with all constants:

(note if you add up the three vectors on the right, you get back to the original solution so it checks). Next we factor z out of the first vector and v out of the second:

This is the general solution in vector form. It can be thought of as having two fundamental segments to it: a homogeneous part:

and a particular part:

We refer to them, in shorthand terms , as x_h and x_p, respectively (the boldface indicating that they are vectors, not single variables). Every possible solution to the system can be represented by x_h + x_p for some values of z and v. That is why it is called the general solution – it covers all special cases.

Why do we call them by these names?

if we simply set v and z = 0, we still have a solution which happens to be

hence it is a particular solution, in the literal sense of the word. If we substituted this back into the original equation, it would work and would satisfy

Ax_p = b (1)

Adding on combinations of

and simply give the solution more generality.

Suppose the original problem had been homogeneous – this would mean the right hand side was all 0s or that b = 0. One would still solve the problem in the same manner only there would be no constant terms in the solution, the 6, 12 and 9 would be gone, and there would only be the terms with z's and v's in them, and they would be the same as what we got. Hence they would solve the problem Ax = 0 and are thus reasonably called "homogeneous solutions". Algebraically,

Ax_h = 0 (2)

We know that matrix multiplication is linear so we can check out the general solution as follows:

A(x_h + x_p) = Ax_h + Ax_p = 0 + b = 0

This simply verifies that it is ok to add on the homogeneous solutions and that you still have a "solution" in the literal sense of the word solution.

Adding on the homogenous solutions has analogues in calculus:

in Integral Calculus, we add a constant C on to an antiderivative to provide a more general answer.

in Differential Equations, we separately solve the homogeneous and particular problems and add the results together to get the general solution. A problem such as

y'' + 25 y = t²

has homogeneous solutions of sin(5t) and cos(5t). It has a particular solution of

y(t) =2t²/25 - 2/625

so the general solution is the homogeneous plus the particular

y(t) = C₁sin(5t) + C₂cos(5t) +2t²/25 - 2/625

(the constants C₁ and C₂ are "arbitrary" like our z and v are earlier). In short, the situation in Differential Equations in completely analogous to the one in Linear Algebra. These sorts of problems are all part of Linear Mathematics.