Engineering mathematics lecture notes, Lecture notes of Engineering Mathematics

Download Engineering mathematics lecture notes and more Lecture notes Engineering Mathematics in PDF only on Docsity!

ENGINEERING MATHEMATICS

KMM 511E

SOLVING LINEAR SYSTEMS OF

EQUATIONS

A linear system with m equations and n unknowns

a

11

x

1

+ a

12

x

2

+ a

13

x

3

+ … = b

1

a

21

x

1

+ a

22

x

2

+ a

23

x

3

+ … = b

2

a

m

x

1

+ a

m

x

2

+ a

m

x

3

+ … = b

m

a and b represent coefficients and x represents unknowns

When all b’s are equal to 0, then the system is homogeneous

We obtain systems of equations when relations between two or more variables are

provided for two or more situations

Augmented matrix ( Ã)

a 11 a 12 b 1 a 21 a 22 b 2

Solving Linear Systems of Equations

Elimination Methods The first step is forming the augmented matrix of the related linear system Then a triangular matrix is formed by eliminative operations These operations may be: Adding a row with another row Multiplication of a row by a scalar value Interchanging two rows (and sometimes columns)  pivoting Finally, the unknowns may be calculated by back-substitution

Gauss Elimination Method

2x 1 + 5 x 2 = 2 4 x 1 + 3 x 2 = 18 A = 2 5 coefficient matrix 4 3 Ã = 2 5 2 augmented matrix 4 3 18 In order to form an upper triangular matrix, we may select the first row as the pivot row and multiply it with - 2, the result will be added to the second row forming a new second row. In this manner, the first element of the second row will be eliminated. 2 5 2 0 - 7 14 From the second row, we conclude that - 7x 2 = 14 and x 2 = - 2 We can back-substitute the result in the first row to obtain 2x 1 + 5(-2) = 2 and x 1 = 6

Gauss-Jordan Elimination Method

2x 1 + 5 x 2 = 2 4 x 1 + 3 x 2 = 18 A = 2 5 coefficient matrix 4 3 Ã = 2 5 2 augmented matrix 4 3 18 Here, we should form a diagonal or unit matrix. If we select the first row as the pivot row and multiply it with - 2, the result will be added to the second row forming a new second row. In this manner, the first element of the second row will be eliminated. We continue to eliminate the second element of the first row. For this purpose, we multiply the second row by 5/7 and add it to the first row. If we divide the first row by 2 and the second row by - 7 we obtain a unit matrix. 2 5 2 2 0 12 1 0 6 0 - 7 14  0 - 7 14  0 1 - 2 From the second row, we conclude that x 2 = - 2 and from the first row we see that x 1 = 6

Linear Systems are

• Consistent if there is at least one solution
• Inconsistent if there is no solution
• Overdetermined if there are extra equations
• Underdetermined if the equations are less than the unknowns
• Determined if the numbers of equations and unknowns are equal A linear system of equations may have
• No solution : parallel lines
• Single solution : intersecting lines
• Infinite number of solutions : coinciding lines We may predict the existence of a solution for the linear systems by calculating the ranks of the coefficient (nxn) and augmented matrices When
• Rank of A  Rank of à then there is no solution
• Rank of A = Rank of à = n then there is a single solution
• Rank of A = Rank of à  n then there is an infinite number of solutions

0 0 - 1 The rank of this 2x3 B matrix is B1 = 2 0 0 - 1 det(B1) = - 2 so rank of B = 2 Example of a single solution system

• x 1 + x 2 + 2x 3 = 2 3x 1 – x 2 + x 3 = 6
• x 1 + 3x 2 + 4x 3 = 4 Example of a no solution system 3x 1 + 2x 2 + x 3 = 3 2x 1 + x 2 + x 3 = 0 6x 1 + 2x 2 + 4x 3 = 6 When there is no solution, there will be an inconsistency for the result obtained at the end of elimination. In one of the rows  0 0 0 12

Example of a system with infinite number of solutions 3x 1 + 2x 2 + 2x 3 - 5x 4 = 8 0.6x 1 + 1.5x 2 + 1.5x 3 – 5.4x 4 = 2. 1.2x 1 – 0.3x 2 – 0.3x 3 + 2.4x 4 = 2. When there are infinite number of solutions, one of the rows will have all 0s In one of the rows  0 0 0 0 Actually, here we have 4 unknowns but 3 equations

For singular matrices, there is no inverse matrix

• • 1 1 2 1 0 0 - 1 1 2 1 0 0 1 0 0 - 0.7 0.2 0.
  • 0 2 7 3 1 0  0 2 7 3 1 0 ….. 0 1 0 - 1.3 - 0.2 0.
  • 0 2 2 - 1 0 1 0 0 - 5 - 4 - 1 1 0 0 1 0.8 0.2 - 0.
    • • 0.7 0.2 0.
• A-^1 = - 1.3 - 0.2 0. - 0.8 0.2 - 0.

Another method of solving linear systems of equations

Cramer’s Rule

a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 Let us multiply the first row by a 22 and the second row by – a 12 a 11 a 22 x 1 + a 12 a 22 x 2 = b 1 a 22

• a 21 a 12 x 1 - a 22 a 12 x 2 = - b 2 a 12 (a 11 a 22 - a 12 a 21 )x 1 = b 1 a 22 - b 2 a 12 x 1 = (b 1 a 22 - b 2 a 12 ) / (a 11 a 22 - a 12 a 21 ) det(A) = a 11 a 12 a 21 a 22 = a 11 a 22 – a 12 a 21 = D D 1 = det b 1 a 12 b 2 a 22 = b 1 a 22 – b 2 a 12

LU Decomposition (Factorization)

This method is about factorizing a matrix as the product of a lower triangular matrix

and an upper triangular matrix

A = L U

where A is a square matrix

A x = L U x = b

There are different versions of this method

Doolittle Method

For a matrix A of 2x

A = 2 3 = 1 0 2 3 = L U

We can form first the upper triangular matrix by elimination and then by considering matrix multiplication obtain the elements of the lower triangular matrix For a square matrix of nxn, we may construct the upper and lower matrices as u 11 u 12 u 13 1 0 0 0 u 22 u 23  m 21 1 0 0 0 u 33 m 31 m 32 1

By assuming that Ux = y  Ly = b

1 0 0 y 1 8

0 1 0 y 2 = - 7

2 - 1 1 y 3 26

Solving the linear system starting from y 1 gives

y = - 7

Ux = y

3 5 2 x 1 8

0 8 2 x 2 = - 7

0 0 6 x 3 3

Solving the linear system starting from x 3 gives

x = - 1