lgebra Linear e Aplicaes RECTANGULAR SYSTEMS AND ECHELON FORMS - - PowerPoint PPT Presentation

Álgebra Linear e Aplicações

RECTANGULAR SYSTEMS AND ECHELON FORMS

Rectangular systems

• General linear system
• What happens to elimination when m ≠ n?

a11x1 + a12x2 + · · · a1nxn = b1 a21x1 + a22x2 + · · · a2nxn = b2 . . . am1x1 + am2x2 + · · · amnxn = bm

More on matrix notation

• Matrix of size m × n

A =      a11 a12 · · · a1n a21 a22 · · · a2n . . . . . . ... . . . am1 am2 · · · amn      = [aij]m×n

m = n: square matrix m n: rectangular matrix

Ai∗ = ai1 ai2 · · · ain

• row vector

A∗j =      a1j a2j . . . amj     

column vector

    1 2 1 3 3 −2 −2 −2 2 2 2 −2 −2 1         1 2 1 3 3 −2 −2 −2 2 2 2 −2 −2 1    

Gaussian Elimination example

• Consider the system
• Gaussian Elimination

x1 + 2x2 + x3 + 3x4 + 3x5 = 5 2x1 + 4x2 + 4x4 + 4x5 = 6 x1 + 2x2 + 3x3 + 5x4 + 5x5 = 9 2x1 + 4x2 + 4x4 + 7x5 = 9     1 2 1 3 3 2 4 4 4 1 2 3 5 5 2 4 4 7         1 2 1 3 3 −2 −2 −2 3         1 2 1 3 3 −2 −2 −2 3    

Modified Gaussian Elimination

• Look for a column that

has a viable pivot

• If needed, exchange

rows to move pivot up

• Stop if all rows below

last pivot are zero

    ∗ ∗ ∗ ∗ ∗ s s s s s s s s s s s s         ∗ ∗ ∗ ∗ ∗ s s s s s s s s s         ∗ ∗ ∗ ∗ ∗ s s ∗ s s s s s         ∗ ∗ ∗ ∗ ∗ ∗ s s s s s s s         ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗    

Row Echelon Form

• Two conditions
• All rows below a zero-row are also zero
• All elements below or left of pivots are zero
• Elements not unique, but form is unique!

        ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗        

Rank of a Matrix

• Let A have row echelon form E
• rank(A) = number of pivots
• rank(A) = number of non-zero rows in E
• rank(A) = number of basic columns in A
• Basic column is a column at a pivotal position

A =     1 2 1 3 3 2 4 4 4 1 2 3 5 5 2 4 4 7     E =     1 2 1 3 3 −2 −2 −2 3     A∗1 =     1 2 1 2     A∗3 =     1 3     A∗5 =     3 4 5 7    

Modified Gauss-Jordan example

• Back to our system

    1 2 1 3 3 −2 −2 −2 3         1 2 1 3 3 1 1 1 3         1 2 2 2 1 1 1 3         1 2 2 2 1 1 1 1         1 2 2 1 1 1    

Reduced Row Echelon Form

• Echelon form is result of Gaussian Elimination
• Gauss-Jordan leads to reduced row echelon form
• Notation: EA is reduced row echelon form of A
• Reduced form is unique (proof later on)
• Useful for theoretical results

        1 ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ 1 ∗ ∗ ∗ 1 ∗        

Why basic columns are basic

• Non-basic columns are combinations of basic

columns, as made obvious by reduced form

• Row operations preserve column relationships

EA =     1 2 2 1 1 1     A =     1 2 1 3 3 2 4 4 4 1 2 3 5 5 2 4 4 7     E∗2 = 2E∗1 E∗4 = 2E∗1 + E∗3 A∗2 = 2A∗1 A∗4 = 2A∗1 + A∗3

Column relationships in A and EA

• are the basic columns to the left of E*k
• Same relationships hold for and A*k
• Why?

E∗k =           µ1 µ2 . . . µj . . .           = µ1           1 . . . . . .           + µ2           1 . . . . . .           + · · · + µj           . . . 1 . . .           = µ1E∗b1 + µ2E∗b2 + · · · + µjE∗bj

A∗bi E∗bi

A∗k = µ1A∗b1 + µ2A∗b2 + · · · + µjA∗bj

SOLVING RECTANGULAR SYSTEMS

Consistency

• Rectangular system is consistent if it has

at least one solution

• It is inconsistent if it has no solution

Consistent system Inconsistent system

Using Gaussian elimination instead

• Reduce [A|b] to [E|c] in row echelon form
• If , we have a problem
• Otherwise, back-substitution works!

         ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ α . . . . . . . . . . . . . . . . . . . . . . . . . . .         

α 6= 0

0x1 + 0x2 + · · · + 0xn = α

Equivalent consistency criteria

• When row reducing [A|b], a row of the

following form never appears

• b is a combination of the basic columns in A
• rank[A|b] = rank(A)

· · · α α 6= 0

Homogeneous systems

• An homogeneous system is of the form
• Can it be inconsistent?
• Values of ?
• Trivial solution!

a11x1 + a12x2 + · · · a1nxn = 0 a21x1 + a22x2 + · · · a2nxn = 0 . . . am1x1 + am2x2 + · · · amnxn = 0

α 6= 0

x1 = x2 = · · · = xm = 0

Homogeneous system example

• Consider the system
• In row echelon form
• Equivalent to
• 4 unknowns and only

2 equations!

• Pick basic variables
• pivotal positions
• as functions of free variables

x1 + 2x2 + 2x3 + 3x4 = 0 2x1 + 4x2 + x3 + 3x4 = 0 3x1 + 6x2 + x3 + 4x4 = 0   1 2 2 3 2 4 1 3 3 6 1 4     1 2 2 3 −3 −3 −5 −5     1 2 2 3 −3 −3   x1 + 2x2 + 2x3 + 3x4 = 0 − 3x3 − 3x4 = 0 x3 = −x4 x1 = −2x2 − 2x3 − 3x4 = −2x2 − x4

Representating the general solution

• Pick any values for x2 and x4, and you create a

new solution to the system

• Every solution is a combination of two

particular solutions

x1 = −2x2 − x4 x2 = x2 x3 = −x4 x4 = x4 h1 =     −2 1     h2 =     −1 −1 1         x1 x2 x3 x4     =     −2x2 − x4 x2 −x4 x4     = x2     −2 1     + x4     −1 −1 1    

General homogeneous system

• Consider an m×n system [A|0]
• How many basic variables?
• r = rank(A)
• How many free variables?
• n – r
• General form of solution is

x = xf1h1 + xf2h2 + · · · + xfn−rhn−r

Unicity of homogeneous solution

• Consistency has already been shown
• Trivial solution
• Look at general solution
• When is it unique?
• When there are no free variables
• When n-r = 0
• i.e., when rank(A) = n

x1 = x2 = · · · = xm = 0 x = xf1h1 + xf2h2 + · · · + xfn−rhn−r

Nonhomogeneous systems

• A system is non-homogeneous
• whenever for at least one i
• Solution follows the same procedure as the

homogeneous case

a11x1 + a12x2 + · · · a1nxn = b1 a21x1 + a22x2 + · · · a2nxn = b2 . . . am1x1 + am2x2 + · · · amnxn = bm

bi 6= 0

Solving nonhomogeneous systems

• Start with [A|b] and reduce to [E|c]
• Identify the basic and free variables
• Apply back-substitution to solve for basic

variables as function of free variables

• Write the result in the form
• The only difference is the term p that was not

needed in the homogeneous case

x = p + xf1h1 + xf2h2 + · · · + xfn−rhn−r

Example showing the p term

• The system
• Augmented matrix
• Gauss-Jordan result
• Reduced system
• Back substitution
• General form

x1 + 2x2 + 2x3 + 3x4 = 4 2x1 + 4x2 + x3 + 3x4 = 5 3x1 + 6x2 + x3 + 4x4 = 7 E[A|b] =   1 2 1 2 1 1 1   [A|b] =   1 2 2 3 4 2 4 1 3 5 3 6 1 4 7   x1 + 2x2 + x4 = 2 x3 + x4 = 1 x1 = 2 − 2x2 − x4 x2 = x2 x3 = 1 − x4 x4 = x4     x1 x2 x3 x4     =     2 1     + x2     −2 1     + x4     −1 −1 1    

General solution of nonhomogeneous system

• Particular solution + general solution of the

associated homogeneous system!

E[A|b] = [EA|c] =            1 ∗ ∗ ∗ ∗ c1 1 ∗ ∗ ∗ c2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 ∗ ∗ ∗ cr−1 1 ∗ cr            xbi = αifixfi + αifi+1xfi+1 + · · · αifn−rxfn−r xbi = ci + αifixfi + αifi+1xfi+1 + · · · αifn−rxfn−r E[A|0] = [EA|0] =            1 ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 ∗ ∗ ∗ 1 ∗           

Summary

• General solution of linear system is
• Column p is a particular solution to

nonhomogeneous system and is the general solution of the associated homogeneous system

• Unique solution if and only if any of
• rank(A) = n = number of unknowns
• There are no free variables
• Associated homogeneous system has only trivial solution

xh = xf1h1 + xf2h2 + · · · + xfn−rhn−r x = p + xh

Application to electric circuits

Ohm’s law and Kirchhoff’s rules

• Reduce problem to linear system
• Ohm’s Law
• For a current I, the voltage drop across a

resistor R is V = R I

• Kirchhoff’s rules
• Nodes: algebraic sum of currents entering a

node is zero

• Loop: algebraic sum of voltage drops along each

loop is zero

Formulating the linear system

• Nodes
• Loops

A : I1R1 − I3R3 + I5R5 = E1 − E3 B : I2R2 − I5R5 + I6R6 = E2 C : I3R3 + I4R4 − I6R6 = E3 + E4 1 : I1 − I2 − I5 = 0 2 : −I1 − I3 + I4 = 0 3 : I3 + I5 + I6 = 0 4 : I2 − I4 − I6 = 0