Cramer's Rule and Gauss Elimination - Welcome — TTU · PDF fileOutlines Part I: Review of Previous Lecture Part II: Cramer’s Rule and Gauss Elimination Review of Previous Lecture

Outlines

Cramer’s Rule and Gauss Elimination

Mike Renfro

September 28, 2004

Mike Renfro Cramer’s Rule and Gauss Elimination

OutlinesPart I: Review of Previous LecturePart II: Cramer’s Rule and Gauss Elimination

Review of Previous Lecture


OutlinesPart I: Review of Previous LecturePart II: Cramer’s Rule and Gauss Elimination


Cramer’s RuleIntroductionExampleAdvantages and Disadvantages

Gauss EliminationIntroduction and RulesExampleMatrix Version and ExampleAdvantages and Disadvantages

Homework


Part I




Graphical interpretation

Solvable and unsolvable problems

Linear dependence and independence

Ill-conditioning


Cramer’s RuleGauss Elimination

Homework

Part II




Homework

IntroductionExampleAdvantages and Disadvantages

Cramer’s Rule (1750)

A linear system of equations can be solved by using Cramer’s rule,which for a system of 2 equations[

a11 a12

a21 a22

]{x1

x2

}=

{b1

b2

}yields

x1 =|[A1]||[A]|

, x2 =|[A2]||[A]|

where

A1 =

[b1 a12

b2 a22

]A2 =

[a11 b1

a21 b2

]



Homework


Cramer’s Rule Details

For a system of n equations, Cramer’s rule requires that youcalculate n + 1 determinants of n × n matrices.

In the general case for a system of equations [A]{x} = {b},the matrix [Ai ] is obtained by replacing the ith column of theoriginal [A] matrix with the contents of the {b} vector.

Each unknown variable xi is found by dividing the determinant|[Ai ]| by the determinant of the original coefficient matrix|[A]|.



Homework


Cramer’s Rule Example

Solve the following system of equations using Cramer’s rule:

x1 − x2 + x3 = 3

2x1 + x2 − x3 = 0

3x1 + 2x2 + 2x3 = 15

Convert the system of equations into matrix form: 1 −1 12 1 −13 2 2

x1

x2

x3

=

30

15



Homework


Cramer’s Rule Example (continued)

[A] =

1 −1 12 1 −13 2 2

, {x} =

x1

x2

x3

, {b} =

30

15

Define matrices [A1], [A2], and [A3] as

[A1] =

3 −1 10 1 −1

15 2 2

, [A2] =

1 3 12 0 −13 15 2

,

[A3] =

1 −1 32 1 03 2 15



Homework



Calculate determinants of [A], [A1], [A2], and [A3]:

|[A]| = 12

|[A1]| = 12

|[A2]| = 24

|[A3]| = 48

Unknowns x1, x2, and x3 are then calculated as

x1 =|[A1]||[A]|

=12

12= 1, x2 =

|[A2]||[A]|

=24

12= 2, x3 =

|[A3]||[A]|

=48

12= 4



Homework



Be sure to double-check your answers by substituting them intothe original equations:

x1 − x2 + x3 = 1 − 2 + 4 = 3

2x1 + x2 − x3 = 2 + 2 − 4 = 0

3x1 + 2x2 + 2x3 = 3 + 4 + 8 = 15



Homework


Cramer’s Rule Advantages/Disadvantages

Advantages

Easy to remember steps

Disadvantages

Computationally intensive compared to other methods: themost efficient ways of calculating the determinant of an n × nmatrix require (n − 1)(n!) operations. So Cramer’s rule wouldrequire (n − 1)((n + 1)!) total operations. For 8 equations,that works out to 7(9!) = 2540160 operations, or around 700hours if you can perform one operation per second.

Roundoff error may become significant on large problems withnon-integer coefficients.



Homework

Introduction and RulesExampleMatrix Version and ExampleAdvantages and Disadvantages

Gauss Elimination

Recall the scaffolding problem from the beginning of Chapter 3. Itsmatrix form was

1 1 −1 −1 0 −10 −9 1 4 0 70 0 1 1 −1 00 0 0 −3 2 00 0 0 0 1 10 0 0 0 0 −4

TA

TB

TC

TD

TE

TF

=

P1

−5P1

P2

−P2

P3

−P3

Notice that its coefficient matrix contains nothing but zeroesbelow the diagonal. This is an example of an upper triangularmatrix, and these systems of equations are very easy to solve.



Homework


Gauss Elimination Introduction (continued)

The original system of equations on the scaffolding problem was

TA +TB −TC −TD −TF = P1

−9TB +TC +4TD +7TF = −5P1

TC +TD −TE = P2

−3TD +2TE = −P2

TE +TF = P3

−4TF = −P3

.

Notice that we can solve for TF using only the sixth equation inthe system. That is, TF = P3

4 . After solving for TF , we can solvefor TE using only the fifth equation. The pattern continues,back-substituting through the system of equations until finally wesolve for TA using the first equation.



Homework


Gauss Elimination Introduction (continued)

The goal of Gauss elimination is to convert any given system ofequations into an equivalent upper triangular form. Onceconverted, we can back-substitute through the equations, solvingfor the unknowns algebraically.



Homework


Gauss Elimination Rules

The operations used in converting a system of equations to uppertriangular form are known as elementary operations and are:

Any equation may be multiplied by a nonzero scalar.

Any equation may be added to (or subtracted from) anotherequation.

The positions of any two equations in the system may beswapped.



Homework


Gauss Elimination Example

2x1 − x2 + x3 = 4 (1)

4x1 + 3x2 − x3 = 6 (2)

3x1 + 2x2 + 2x3 = 15 (3)

To eliminate the 4x1 term in Equation 2, multiply Equation 1 by 2and subtract it from Equation 2. To eliminate the 3x1 term inEquation 3, multiply Equation 1 by 3

2 and subtract it fromEquation 3. This gives a system of equations

2x1 − x2 + x3 = 4 (4)

5x2 − 3x3 = −2 (5)

7

2x2 +

1

2x3 = 9 (6)



Homework


Gauss Elimination Example (continued)

To eliminate the 72x2 term from Equation 6, multiply Equation 5

by 710 and subtract it from Equation 6. This gives a system of

equations

2x1 − x2 + x3 = 4 (7)

5x2 − 3x3 = −2 (8)

13

5x3 =

52

5(9)



Homework


Gauss Elimination Example (continued)

Equation 9 can easily be solved for x3.

x3 =52

5

(5

13

)= 4

Equation 8 can easily be solved for x2, once x3 is known.

x2 =1

5(−2 + 3x3) =

1

5(−2 + 3(4)) = 2

Equation 7 can easily be solved for x1, once both x2 and x3 areknown.

x1 =1

2(4 + x2 − x3) =

1

2(4 + 2 − 4) = 1



Homework


Matrix Version of Gauss Elimination

The Gauss elimination method can be applied to a system ofequations in matrix form. Instead of eliminating terms fromequations, we’ll be replacing certain elements of the coefficientmatrix with zeroes.



Homework


Matrix Version (Step 0)

Start by defining the augmented matrix [C (0)] for the problem:

[C (0)] =

a(0)11 a

(0)12 · · · a

(0)1n a

(0)1,n+1

a(0)21 a

(0)22 · · · a

(0)2n a

(0)2,n+1

......

......

...

a(0)n1 a

(0)n2 · · · a

(0)nn a

(0)n,n+1

where the first n columns are the elements of the original [A]matrix, and the last column is the elements of the original {b}matrix.



Homework



Zero out the first column of the [C ] matrix, rows 2 · · · n. To turna21 to a zero, multiply row 1 by a21

a11, then subtract the numbers on

row 1 from row 2. To turn a31 to a zero, multiply row 1 by a31a11

,then subtract the numbers on row 1 from row 3. Repeat for rows4 · · · n.

[C (1)] =

a(0)11 a

(0)12 · · · a

(0)1n a

(0)1,n+1

0 a(1)22 · · · a

(1)2n a

(1)2,n+1

......

......

...

0 a(1)n2 · · · a

(1)nn a

(1)n,n+1



Homework



Zero out the second column of the [C ] matrix, rows 3 · · · n. Toturn a32 to a zero, multiply row 2 by a32

a22, then subtract the

numbers on row 2 from row 3. To turn a42 to a zero, multiply row2 by a42

a22, then subtract the numbers on row 2 from row 4. Repeat

for rows 5 · · · n.

[C (1)] =

a(0)11 a

(0)12 a

(0)13 · · · a

(0)1n a

(0)1,n+1

0 a(1)22 a

(1)13 · · · a

(1)2n a

(1)2,n+1

0 0 a(2)33 · · · a

(2)3n a

(2)3,n+1

......

......

......

0 0 a(2)n3 · · · a

(2)nn a

(2)n,n+1



Homework


Matrix Version (Step n-1)

Zero out the (n − 1)th column of the [C ] matrix, row n. To turnan,n−1 to a zero, multiply row n − 1 by

an,n−1

an−1,n−1, then subtract the

numbers on row n − 1 from row n.

[C (1)] =

a(0)11 a

(0)12 a

(0)13 · · · a

(0)1n a

(0)1,n+1

0 a(1)22 a

(1)13 · · · a

(1)2n a

(1)2,n+1

0 0 a(2)33 · · · a

(2)3n a

(2)3,n+1

......

......

......

0 0 0 · · · a(n−1)nn a

(n−1)n,n+1



Homework


Example

Solve the following system of equations with Gauss elimination: 2 −1 14 3 −13 2 2

x1

x2

x3

=

46

15

First, set up the augmented matrix [C (0)]:

[C (0)] =

2 −1 1 44 3 −1 63 2 2 15



Homework


Example (continued)

Step 1a: eliminate the 4 on row 2, column 1. Multiply all theelements of row 1 by a21

a11= 4

2 = 2, then subtract them from theelements of row 2.

[C ] =

2 −1 1 44 − (2)(2) 3 − (2)(−1) −1 − (2)(1) 6 − (2)(4)

3 2 2 15

=

2 −1 1 40 5 −3 −23 2 2 15



Homework


Example (continued)

Step 1b: eliminate the 3 on row 3, column 1. Multiply all theelements of row 1 by a31

a11= 3

2 = 1.5, then subtract them from theelements of row 3.

[C (1)] =

2 −1 1 40 5 −3 −2

3 − (1.5)(2) 2 − (1.5)(−1) 2 − (1.5)(1) 15 − (1.5)(4)

=

2 −1 1 40 5 −3 −20 3.5 0.5 9

This completes the first elimination step.



Homework


Example (continued)

Step 2: eliminate the 3.5 on row 3, column 2. Multiply all theelements of row 2 by a32

a22= 3.5

5 = 0.7, then subtract them from theelements of row 3.

[C (2)] =

2 −1 1 40 5 −3 −20 3.5 − (0.7)(5) 0.5 − (0.7)(−3) 9 − (0.7)(−2)

=

2 −1 1 40 5 −3 −20 0 2.6 10.4

This completes the second elimination step.



Homework


Example (continued)

We’ve now converted the original system of equations 2 −1 14 3 −13 2 2

x1

x2

x3

=

46

15

into an equivalent upper-triangular system of equations 2 −1 1

0 5 −30 0 2.6

x1

x2

x3

=

4

−210.4



Homework


Example (continued)

The new system of equations can be converted back to algebraicform as:

2x1 − x2 + x3 = 4 (10)

5x2 − 3x3 = −2 (11)

2.6x3 = 10.4 (12)

Solve Equation 12 for x3: x3 = 10.42.6 = 4. Then solve Equation 11

for x2: x2 = 15(−2 + 3x3) = 2. Then solve Equation 10 for x1:

x1 = 12(4 + x2 − x3) = 1.



Homework


Example (continued)

Double-check the solution by substituting the values of x1, x2, andx3 into the original equations:

2x1 − x2 + x3 = 2(1) − 2 + 4 = 4

4x1 − 3x2 − x3 = 4(1) + 3(2) − 4 = 6

3x1 + 2x2 + 2x3 = 3(1) + 2(2) + 2(4) = 15



Homework


Gauss Elimination Advantages/Disadvantages

Advantages

Much less computation required for larger problems. Gausselimination requires n3

3 multiplications to solve a system of nequations. For 8 equations, this works out to around 170operations, versus the roughly 2.5 million operations forCramer’s rule.

Disadvantages

Not quite as easy to remember the procedure for handsolutions.

Roundoff error may become significant, but can be partiallymitigated by using more advanced techniques such as pivotingor scaling.



Homework

Homework

Solve Problem 3.4 using: Cramer’s rule, Gauss elimination,and MATLAB’s \ operator. Double-check your answers bysubstituting them back into the original equations.


Documents

Cramer's Rule and Gauss Elimination - Welcome — TTU · PDF fileOutlines Part I: Review of Previous Lecture Part II: Cramer’s Rule and Gauss Elimination Review of Previous Lecture