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Lecture 6

Linear Algebra (III)

Linear Systems of Equations

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Linear System of Equations

• A system of m-linear equations in n variables x1, x2, ..., xn has the general form

(1)

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Linear System of Equations

• where the coefficients aij(i = 1, 2, ...,m; j = 1, 2, ..., n)

• and the quantities bi

• are all known scalars (numbers).

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This is a Linear System

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This is not a Linear System

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Matrix Representation of a Linear System of Equations

• Any linear system of the form (1) can be written in the matrix form

AX = B

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Matrix Representation of a Linear System of Equations

• With

A is the coefficient matrix

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Matrix Representation of a Linear System of Equations

X is the column of variables

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Matrix Representation of a Linear System of Equations

B is the column of constants

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Example

• Can be represented as

35

12

654

132

z

y

x

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Example• The matrix form

• Represents the system

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Gauss Elimination Method

• The method consists of four steps 1. Construct an augmented matrix for the

given system of equations. 2. Use elementary row operations to

transform the augmented matrix into an augmented matrix in row-reduced form.

3. Write the equations associated with the resulting augmented matrix.

4. Solve the new set of equations by back substitution.

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Augmented Matrix

• The augmented matrix for AX = B is the partitioned matrix [A|B]

• E.g.

• has its augmented matrix as

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Elementary Row Operationselementary row operations are :

• 1- Interchange any two rows in a matrix

• 2- Multiply any row of a matrix by a nonzero scalar

• 3- Add to one row of a matrix a scalar times another row of the same matrix

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Row-Reduced Form• A matrix is in row-reduced form if it satisfies the

following four conditions:

1. All zero rows appear below nonzero rows when both types are present in the matrix.

2. The first nonzero element in any nonzero row is 1.

3. All elements directly below (that is, in the same column but in succeeding rows form) the first (left- to- right)

nonzero element of a nonzero row are 0 .

4. The first nonzero element of any nonzero row appears in a later column (further to right) than the first nonzero

element in any preceding row.

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Example• Use Gaussian elimination to solve the

system

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Step 1The augmented matrix

• The augmented matrix of the system is:

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Step 2Elementary row operations

We use elementary row operations to transform the augmented matrix into

row-reduced form as follows,

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This is a Row Reduced Form

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Step 3The Resulting System• We write the equations of the resulting

augmented matrix

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Step 3Writing the Resulting System

• We write the equations associated with the resulting augmented matrix

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Step 4Solving the Resulting System

• we Solve the derived set of equations by back substitution.

• The third equation implies that z = 5,

• Substituting in the second equation, we get y = 12 − 15 = −3,

• substituting with the values of z and y in the first equation, we get x = 4

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Example

• Use Gauss elimination method to solve the linear system of equations

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Example

100

010

001

10

01

3

2

I

I

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The Augmented Matrix

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Getting Row Reduced Form

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Getting Row Reduced Form

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Writing the Resulting System

• The resulting system of equations is

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Solving the Equations

• Solving the last system by back substitution, we get the solution

• x = 1 and y = 1

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Example

• Use Gauss elimination method to solve the linear system of equations

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The Augmented Matrix

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Getting Row Reduced Form

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Getting Row Reduced Form

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Writing the Resulting System

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Important Note

• Since the final system has the number of variables (4) greater than the number of equations (3),

• then one of the variable will be arbitrary,

• and the other variables will be found in terms of it.

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Solving The Resulting System

• The solution will be found in terms of x4 as follows,

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Note

• Since x4 is arbitrary, then the system has infinite number of solutions, depending on the value of x4

• for example if you choose

• Then,

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Japan !

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