Page 43 - Numerical Methods for Chemical Engineering

32 1 Linear algebra

N
and use a similar expansion for our specific b ∈ in Ax = b,
b = b 1 u [1] +· · · + b P u [P] + b P+1 u [P+1] +· · · + b N u [N] (1.163)

We now write Av = b to consider when a solution v = x exists,
[1] [P] [P+1] [N]
A v 1 u + ··· + v P u + A v P+1 u + ··· + v N u
= b 1 u [1] +· · · + b P u [P] + b P+1 u [P+1] +· · · + b N u [N] (1.164)

Noting that v 1 u [1] +· · · + v P u [P] ∈ K A ,wehave

[P+1] [N]
A v P+1 u + ··· + v N u
= b 1 u [1] +· · · + b P u [P] + b P+1 u [P+1] +· · · + b N u [N] (1.165)

The vector Avon the left-hand side must be in R A , and thus must have no nonzero component
in K A ; i.e., A[v P+1 u [P+1] +· · · + v N u [N] ] · w = 0 for any w ∈ K A . If any of the coeffi-
cients b 1 ,..., b P on the right-hand side are nonzero, the two sides of the equation cannot
agree. Therefore, a solution to Ax = b exists if for every w ∈ K A , b · w = 0. However, if
the null space is not empty, this solution cannot be unique, because for any w ∈ K A ,we
also have

A(x + w) = Ax + Aw = b + 0 = b (1.166)
A system Ax = b whose matrix has a nonempty kernel either has no solution, or an infinite
number of them.
We have now identified when the linear system Ax = b will have exactly one solution,
no solution, or an infinite number of solutions; however, these conditions are rather abstract.
Later, in our discussion of eigenvalue analysis, we see how to implement these conditions
for specific matrices A and vectors b.
Here, we have introduced some rather abstract concepts (vector spaces, linear trans-
formations) to analyze the properties of linear algebraic systems. For a fuller theoretical
treatment of these concepts, and their extension to include systems of differential equations,
consult Naylor & Sell (1982).

The determinant

IntheprevioussectionwehavefoundthatthenullspaceofAisveryimportantindetermining
whether Ax = b has a unique solution. For a single equation, ax = b, it is easy to find
whether the null space is empty. If a = 0, the equation has a unique solution x = b/a.If
a = 0, there is no solution if b = 0 and an infinite number if b = 0.
We would like to determine similarly whether Ax = b has a unique solution for N ≥ 1.
We thus define the determinant of A as

c = 0, if K A = 0
det(A) =|A|= (1.167)
0, if ∃w ∈ K A , w = 0