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Systems of Linear Algebraic Equations

الكلية كلية العلوم للبنات     القسم قسم فيزياء الليزر     المرحلة 3
أستاذ المادة نزار سالم شنان الزبيدي       4/27/2011 6:45:32 AM

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

Chapter 1

Systems of Linear Algebraic Equations

1.1 Introduction:

The static mechanical spring-mass system illustrated in figure 1.1

consists of three masses m1 to m3, having weight W1 to W3,

interconnected by five linear springs K1 to K5.

Figure 1.1 Static mechanical spring-mass systems.

When the supporting forces F1 to F3 are removed, the masses move

downward and reach a new static equilibrium configuration, denoted by

x1, x2 and x3 where x1, x2 and x3 measured from original location of the

corresponding masses.

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

Free-body diagrams of the three masses are presented at the bottom of

figure 1.1 performing a static force balance on the three masses yields the

following system of three linear algebraic equation:

(K1 +K2+K3)x1 - K2x2 - K3 x3 = W1 (1.1a)

- K2x1 + (K2+K4)x2 K4x3 = W2 (1.1b)

- K3x1 K4x2 + (K3+K4+K5)x3 = W3 (1.1c)

The static mechanical spring-mass system illustrated in figure 1.1 is used

as the example problem in this chapter to illustrated methods for solving

systems of linear algebraic equations.

Let

K1 = 40 N/cm

K2 = K3 = K4 = 20 N/cm

K5 = 90 N/cm

W1 = W2 = W3 = 20 N

For these values, equation (1.1) becomes:

80x1 20x2 20x3 = 20 (1.2a)

- 20x1 + 40x2 20x3 = 20 (1.2b)

- 20x1 - 20x2 + 130x3 = 20 (1.2c)

The solution to equation (1.2) is

x1= 0.6cm, x2=1.0cm, x3=0.4cm

The equations may be linear or nonlinear chapter1 is devoted to the

solution of systems of linear algebraic equations of the following form:

a11x1 + a12x2 + a13x3 + a1nxn = b1 (1.3a)

a21x1 + a22x2 + a23x3 + a2nxn = b2 (1.3b)

.

an1x1 + an2x2 + an3x3 + annxn = bn (1.3n)

The number of equations can ranged from two to hundreds,

thousands and even millions.

We can call the equations ranged like equation (1.3) a

MATRIX.

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

1.2 Properties of Matrices and Determinants.

1.2.1 Matrix Definitions.

A matrix is a rectangular array of elements (either number or

symbols) which are arranged in orderly rows and columns. Each element

of the matrix is distinct and separate. The location of an element in the

matrix is important. Elements of matrix are generally identified by a

double subscripted lowercase letter, for example, aij where the first

subscript i identifies the row of matrix and second subscript j identifies

the column of the matrix.

( 1,2,......., ; 1,2,.........., )___________(1.5)

..........

.....................

..........

..........

1 2

21 22 2

11 12 1

i n j m

a a a

a a a

a a a

A a

n n nm

m

m

ij

Comparing tow equation (1,3) and (1.5) shows that the coefficients of a

system of linear algebraic equations from the elements of an nxn matrix.

Vectors are a special type of matrix which has only one column or one

row.

A column vector is an nx1 matrix. Thus,

( 1,2,......., ) __________ _________( 1 .6 )

...

[ ] 2

1

i n a

x

x

x

x x

n

i

A row vector is a 1xm matrix, for example

Y = [yj] = [y1 y2 ..ym] (j=1, 2, .m) ____(1.6b)

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

1.2.2 Matrix Algebra

Matrix algebra consists of matrix addition, matrix subtraction and

matrix multiplication. Matrix division is not defined. An analogous

operation is accomplished using the matrix inverse.

Let A and B be two matrices of equal size. Thus

A+B = [aij] + [bij] = [aij +bij] = [Cij] = C ____________(1.7a)

A - B = [aij] - [bij] = [aij - bij] = [Cij] = C ____________(1.7b)

Unequal size matrices cannot be added or subtracted.

A + (B+C) = (A+B) + C _____________(1.8)

A + B = B + A ____________(1.9)

Example 1:

add the two 3x3 matrices A and B to obtain the 3x3 matrix C, where

2 3 1

4 1 2

3 2 1

1 4 3

2 1 4

1 2 3

A and B

Solution:

From Eq. (1.7)

cij = aij + bij

c11 = a11 + b11 = 1 + 3 = 4

c12 = a12 + b12 = 2+2 = 4, etc

A B C

3 7 2

2 2 6

4 4 4

(1 2) (4 3) (3 1)

(2 4) (1 1) (4 2)

(1 3) (2 2) (3 1)

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

Matrix multiplication consists of row-element to column-element

multiplication and summation of the resulting products.

If the size of matrix A is n x m and the size of matrix B is m x r then

( 1,2,..., , 1,2,..., )

1

AB a b c C c a b i n j r

m

k

ij ij ij ij ik kj __(1.10)

Multiplication of the matrix A by the scalar consists of multiplying

each element of A by . Thus

A = [aij] = [ aij] = [bij] = B ______________(1.11)

Example 2

Multiply the 3x3 matrix A and the 3x2 matrix B to obtain the 3x2

matrix C, where

2 1

1 2

2 1

1 4 3

2 1 4

1 2 3

A and B

Solution:

c11 = a11b11+a12b21+a13b31 = (1)(2)+(2)(1)+(3)(2) = 10

c12 = a11b12+a12b22+a13b32 = (1)(1)+(2)(2)+(3)(1) = 8

..

.

.

c32 = a31b12+a32b22+a33b32 = (1)(1)+(4)(2)+(3)(1) = 12

Thus

12 12

13 8

10 8

[ ] ij C c

Systems of Linear Algebraic Equations

Nizar Salim / Lecture 2

Transpose of matrix (AT ) is obtained by inter changing the rows and the

columns of A.

If

m mn

n

a a

a

a a a

A

. . .

. . . . .

. . . . .

. . . .

. .

1

21

11 12 1

then

n nm

m

T

a a

a

a a a

A

. . .

. . . . .

. . . . .

. . . .

. .

1

12

11 21 1

Example 3: transpose matrix A

4 0 0

5 8 1

A

Solution

1 0

8 0

5 4

AT

Example 4: transpose matrix B where

1

2

7

B

Solution: BT 7 2 1

Property: (AB)T = AT BT

The unit matrix is

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

I

Where I = IT

Note: if we consider the two square matrices A and B and AB=I, then B

is the inverse of A which is denoted as A-1. Matrix commute on

multiplication. Thus.

A A-1 =A-1 A = I

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