10.34, Numerical Methods Applied to Chemical Engineering
Professor William H. Green
Lecture #3: Matrix Factorization. Modularization.
v v
U⋅x = v
⎛•
⎜
⎜0
⎜0
⎜
⎜0
⎝
•
•
0
0
•
•
•
0
• ⎞⎛ x1 ⎞ ⎛ v1 ⎞
⎟⎜ ⎟ ⎜ ⎟
• ⎟⎜ x 2 ⎟ ⎜ v 2 ⎟
=
• ⎟⎜ x3 ⎟ ⎜ v3 ⎟
⎟⎜ ⎟ ⎜ ⎟
• ⎟⎠⎜⎝ x 4 ⎟⎠ ⎜⎝ v 4 ⎟⎠
⎛•⎞
⎛•⎞
⎛•⎞
⎛•⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜•⎟ v
⎜•⎟
⎜•⎟
⎜ 0⎟
x1 ⎜ ⎟ + x 2 ⎜ ⎟ + x3 ⎜ ⎟ + x 4 ⎜ ⎟ = (v )
0
0
•
•
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜•⎟
⎜ 0⎟
⎜ 0⎟
⎜ 0⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
function x = backsub(U,v)
N = length(v);
% dimension of matrix
for i = 1:(N-1)
m = N+1-i;
% backwards from bottom matrix
x(m) = v(m)/U(m,m);
% solve equation with one unknown
v = v – x(m)*U(:,m); % move terms involving known x(m) to r.h.s.
end
x(1) = v(1)/U(1,1);
% tested using: U = [1 2 3; 0 4 5; 0 0 6], v = [0 1 3]
% x = [-.75 -.375 .5]
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Gaussian Elimination
⎛ 1 2 3⎞
⎛ 0⎞
⎜
⎟ v ⎜ ⎟
⎜ 4 5 6⎟ * x = ⎜1⎟
⎜ 7 8 0⎟
⎜ 3⎟
⎝
⎠
⎝ ⎠
⎛ 1 2 3 0⎞
⎜
⎟ add - 4 *1st row
Augment : ⎜ 4 5 6 1 ⎟
nd
⎜ 7 8 0 3 ⎟ to 2 row
⎝
⎠
⎛1 2
3 0⎞
⎜
⎟add 2 * 2 nd row
⎜0 - 3 - 6 1⎟
rd
⎜ 0 - 6 - 21 3 ⎟ to 3 row
⎝
⎠
⎛1 2 3 0⎞
⎜
⎟
⎜ 0 - 3 - 6 1 ⎟ ⇒ Upper Triangular solution
⎜0 0 - 9 1⎟
⎝
⎠
*gauss.m*
function [U,v] = gauss(A,b)
% uses Gaussian elimination to convert Ax=b to Ux=v
% where U is upper triangular and A is NxN matrix and b is Nx1 matrix
M = [A b]; N = length(b);
for i = 1:(N-1)
for j = (i+1):N
lambda = -M(j,i)/M(i,i);
M(j,i) = M(j,i)+lambda*M(i,i);
end
end
U = M(i,1:N)
unpacks augmented solution
v = M(i,N+1)
Gaussian Elimination has N3 operations. Not good when N is large (106). Backsub
requires N2 operations.
Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical
Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of
Technology. Downloaded on [DD Month YYYY].
v
a
can span entire 2D vector space (linear combination)
v
b
Figure 1. 2 vectors with a common origin and different directions.
r
v
v
x = c1a + c2b
r
v
v
v
x = c1 a + c 2 b + c3 d
can span entire 3D vector space (linear combination)
v
v
r
*Not linearly independent if: d = c1 a + c 2 b
same direction - not independent
v
a
v
b
Figure 2. 2 vectors with a
common origin and direction.
rank (m) = # of linearly independent column vectors = dimension of space spanned
by column vectors
Existence
A*x=b
solution exists if:
rank(A) == rank ([A b])
⎛ col ⎞
⎛ col ⎞
⎛ col ⎞ v
x1 ⎜⎜ ⎟⎟ + x2 ⎜⎜ ⎟⎟ + L + x N ⎜⎜ ⎟⎟ = (b )
⎝ 1 ⎠
⎝ 2 ⎠
⎝N⎠
r
if rank(A) = N and A is NxN: does not matter what b
is; always a solution
Ay
=
0
If null space of A does not include any
Ax = b
vector beyond 0, then solution unique
A(x+y) =
The null space is the set of all vectors
b
of such that x = 0.
Run *gauss.m* => Error!
NaN: error – not a number (divided by 0)
Need to pivot rows to avoid 0 in diagonal
* see “gausselim_pivot.m” online *
A*x1 = b1
A*x2 = b2
Gauss elimination has nothing
to do with b, only A
L-1 * A * x1 = L-1 * b1
U
LUx = b
Lv = b
L: lower triangular
Alternative method
to Gaussian elimination
v
For more info: help lu
Ux=v
MATLAB: [L,U,P] = lu(A)
A = P’*L*U
(factorizing A)
PAx = Pb
LU
LUx = Pb
Lv = P*b (forward sub)
Ux = v
(backward sub)
10.34, Numerical Methods Applied to Chemical Engineering
Prof. William Green
Lecture 1
Page 2 of 2
Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical
Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of
Technology. Downloaded on [DD Month YYYY].