systems of linear equations 1daniel baur / numerical methods for chemical engineers / linear...

1Daniel Baur / Numerical Methods for Chemical Engineers / Linear Equation Systems

Systems of Linear Equations

x bA

Daniel Baur

ETH Zurich, Institut für Chemie- und Bioingenieurwissenschaften

ETH Hönggerberg / HCI F128 – Zürich

E-Mail: [email protected]

http://www.morbidelli-group.ethz.ch/education/index


Problem Definition

We want to solve the problem

where x is the vector of unknowns, while A and b are given

Assumptions1. The number of equations is equal to the number of unknowns,

therefore A is a square matrix2. All components of A, b and x are real3. A solution exists and it is unique

( , ) ( ,1) ( ,1)n n x n b n A

• A-1 exists

• A is not singular

• A’s columns are linearly independent

• A’s rows are linearly independent

• det(A) is non-zero

• rank(A) is equal to n

• Ax = 0 only if x is a null vector


Analytical Approach

Cramer’s rule (1750):The solution is given by

where Ai is defined as follows

det( )

det( )i

ix A

A

1,1 1,2 1, 1 1 1, 1 1,

2,1 2,2 2, 1 2 2, 1 2,

,1 ,2 , 1 , 1 ,

i i n

i i ni

n n n i n n i n n

a a a b a a

a a a b a a

a a a b a a

A

b replaces the ith column


Calculation of the Determinant

The Laplace formula (1772) allows the computation of the determinant of a square matrix:

where Ci,j is the determinant of the sub-matrix obtained by removing the ith row and the jth column of the matrix, multiplied by (-1)i+j:

1, 1,1

det( )N

i ii

a C

A

, ,( 1) det( )i ji j i jC M 1,1 1, 1 1, 1 1,

1,1 1, 1 1, 1 1,

,1,1 1, 1 1, 1 1,

,1 , 1 , 1 ,

j j n

i i j i j i n

i ji i j i j i n

n n j n j n n

a a a a

a a a a

a a a a

a a a a

M

no jth column

no ith row


A First Numerical Approach: Gauss Elimination Method

The following operations do not change the result:1. Multiply a line by a constant2. Substitute a line with a linear combination of multiple lines3. Permute the order of lines

This can be used to produce a triangular matrix, which allows the solution to be found easily by substitution

Example system

1

2

3

3 2 1 1

1 2 2 5

1 1 1 2

x

x

x


Gauss Elimination Example

• Multiply by -3

• Sum it to 1st line• Multiply by -3

• Sum it to 1st line

• Multiply by -4

• Sum it to 2nd line

Triangular System


Generalization of the Gauss Elimination Method

We want to find a general procedure to replace one entry in the matrix with a zero; for this we define a multiplier l21

Note that

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

A

( ) ( )21 21 21 11( ) ( )21

21 22 22 21 1211 ( ) ( )

23 23 21 13

0new old

new old

new old

a a l aa

l a a l aa

a a l a

11 12 13 11 12 13( ) ( )

21 21 22 23 22 23

31 32 33 31 32 33

1 0 0

1 0 0

0 0 1

new new

a a a a a a

l a a a a a

a a a a a a


The Gauss Elimination in Matrix Notation

If we put all multipliers for one column in one matrix, we get

where

This way, a triangular matrix is easily obtained

(0) (0) (0) (0) (0) (0)11 12 13 11 12 13

(0) (0) (0) (0) (0) (0) (1) (1) (1)21 21 22 23 22 23(0) (0) (0) (0) (1) (1)31 31 32 33 32 33

1 0 0

1 0 0

0 1 0

a a a a a a

l a a a a a

l a a a a a

M A A

( )( )

( )

kijk

ij kjj

al

a

( 1) ( 2) (1) (0) (0)( 1) ( 1)

( 1) ( 2) (1) (0) (0)

n nn n

n nx b

b b

A M M M AA

M M M


Example: Gauss Elimination in Matrix Notation

n(n-1) operations (flops)(n-1)(n-2) operations (flops)

Total number of operations required31 1

2

2 2( 1)

3

n n

j j

nj j j


Gauss Elimination / Transformation Method

The Gauss elimination method is relatively easy to implement (even by hand), but has some distinct disadvantages; namely it Changes the matrix A Requires and changes the coefficient vector b Must be rerun if the vector b changes

If we consider the Gauss method in matrix form on the other hand, we can see that we can use

to transform A and b; M is therefore called the Gauss transformation matrix

( 2) (1) (0)nM M M M


The Gauss Transformation Method

We have so far The final matrix A, which is an right triangular matrix The matrix M, which is a left triangular matrix

The inverse of M is also a left triangular matrix

1 (0)21(0) (1)31 31

1 0 0

1 0

1

l

l l

L M


LR (LU) Factorization

We define our (right triangular) solution matrix as follows

If we multiply with L = M-1 from the left on both sides

( 2) (0) (0)n A MA R MA

1 (0) (0) LR M MA A


LR Factorization (Continued)

The starting matrix A is transformed (factorized) as

If we apply this to a linear equation system, we get

This approach rids us of the disadvantages discussed earlier, because For every vector b, two simple triangular systems must be solved

without factorizing again The matrices L and R can be stored using the elements of A If A is modified, it is often possible to modify L and R accordingly

without re-factorizing Note that Gauss elimination is still needed once to compute L and R

A LR

y bx x b

x y

LA LR

R


Pivoting It is easy to see from the definition of the multiplication

factors, that the diagonal elements at each step (the pivot values) cannot be equal to zero

This is circumvented by reordering the rows of the matrix A by multiplication with a permutation matrix P

This approach is referred to as LRP-factorization(or LR-factorization with partial pivoting)

Example:

a11 = 0 switch the lines x1 = x2 = 1

1 PA PM R LR

2

1 2

1

0.4 0.3 0.1

x

x x


Pivoting (Continued)

Pivoting must also take into account scaling problems;Let us consider a similar example

Pivot elements should have large absolute values

201 2

1 2

2 10 1

0.4 0.3 0.1

x x

x x


How does Matlab do it?

When using left divide \, Matlab chooses a procedure depending on the properties of the problem, i.e. if A is1. Diagonal, x is computed directly by division2. Sparse, square and banded, then banded solvers are

used; Either Gauss elimination without pivoting or LR factorization3. Left or right triangular, then backsubstitution is used4. A permutation of a triangular matrix, it is permuted and 3.

applies5. Symmetric or Hermitian, then a Cholesky factorization is

attempted (A = RR*, where R* is the conjugate transpose of R);

If it fails, another indefinite symmetric factorization is attempted

6. Square but 1 through 5 do not apply, then LR factorization with

partial pivoting is applied7. Not square, then Householder reflections are used to

compute afactorization which leads to a least-squares solution, i.e. a

vector xwhich minimizes the length of Ax – b


Assignment 11. Find online the template for an LR-factorization of a matrix

with partial pivoting. Make the function operational by adding the lines that calculate the M matrix of the current step and the new A matrix. Why does the transformation matrix T appear in the formula for M?

Explain by comparing to the definition of M in the non-pivoting case.

2. Use the function to factorize the following matrix

Test if the factorization worked, i.e. if LR = A. Is L in the form you would expect it to be? What implications does

this have for its application in solving a linear system (see slide 13) and how could you correct for it?

1 2 3

9 8 1

5 0 1

A


Hints


Assignment 21. Create a random matrix A with dimensions 4x4 and a

random column vector b with size 4x1. Solve the system Ax = b.

2. Create a function that computes the determinant of square matrices using the Laplace formula. Use a recursive approach (see hints).

3. Use this function to compute the solution of the linear system above using Cramer’s rule.

4. Do the same for linear equation systems with sizes ranging from 5 to 9.

5. Read out the CPU time required to solve all these systems with both methods (Cramer’s method and A\b) and compare them.


Hints

If you remember the Laplace formula as a sum

you can see that the calculation of the determinant requires the calculation of a determinant. You could let the function call itself to do that (recursive function). Remember that the determinant of a 1x1 matrix is equal to its only element.

There are several Matlab commands to read out timings, however the most reliable one is t_start = tic;statements;t_elapsed = toc(t_start);

Where the time elapsed (in second) is stored in t_elapsed

11, 1, 1, 1,

1 1

det( ) ( 1) det( )N N

ii i i i

i i

a C a

A M

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