| 
  • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today!

View
 

Chapter 2 Linear Algebra (continued)

Page history last edited by sidjaggi 13 years, 7 months ago

Linear Algebra (continued)

 

1 Systems of Linear Equations

1.1 General Form
A linear equation in the n variables or unknows Formula is an equation of the form

                                             Formula

where Formula and Formula are real constants.

This equation can also be expressed as an inner product of two vectors Formula and Formula, i.e. Formula, where

                    Formula

A system of linear equations or a linear system is a finite set of linear equations. For example, a linear system of m equations in n unknowns Formula has the form

                                   Formula

where Formula and Formula are constans.

Note.

     (a)  The Formula are called the Formula of the system.

     (b)  If all Formula are zero, then the system is called a homogeneous system.

     (c)  If at least one Formula is not zero, then the system is called a nonhomogenous system.

     (d)  A solution of the system is a set of numbers Formula that satisfies all the m equations.

 

Example 1.1. A system of two equations in three unknowns is

                                                            Formula

                                                            Formula

The system has a solution Formula, FormulaFormula since these values satisfy both equations. However, Formula, FormulaFormula is not a solution since these values satisfy only the first of the two equations of the system.

 

1.2 Existence of Solution
Every system of linear equations has either     

     (a) no solutions
     (b) exactly one solution
     (c) or infinitely many solutions

The following figure illustrates the three possible solutions for a system of two equation in two unknown Formula.

 

A system of equations that has no solutions is said to be inconsistent, since the system has contradictory equations, e.g.

                                                                 Formula

 

A system of equations that has at least one solution is said to be consistent.

 

1.3 Matrix Form of the Linear System

From the definition of matrix multiplication we can write (2) as

                                                                     Formula

where the coefficient matrix A is the Formula matrix

               Formula

 

are column vectors. Note that x has n components, whereas b has m components.

The matrix

                                             Formula

is called the augmented matrix of the system. The augmented matrix deteremines the system completely because it contains all the given numbers appearing in (2).

 

Example 1.3. The augmented matrix for the system of equations

                                                  Formula 

is

                                                       Formula

 

Exercise 1.3. Write the augmented matrix for the following system of equations

                                                  Formula

 

2 Gaussian Elimination
2.1 Elementary Row Operation

The basis method for solving a system of linear equations is to replace the given system by a new system that has the same solution set but is easier to solve.

This new system is obtained in a series of step by applyin elementary row operations to eliminate unknowns systematically.

     Formula

 

The aim of the elementary row opertions is to reduce the argmented matrix to an echelon or triangular form that has the following properties

     (a) If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix.
     (b) In any two successive rows that do not consist entirely of zeros, the leading non-zero in the low row occurs farther to the right than the leading non-zero in the higher row.

 

Example 2.1.

               Formula

where any real numbers substituted for the *'s.

 

2.2 Gauss Elimination and Back Substitution Method

The Gauss elimination method reduces a system to its echelon form. From this form we can then readily obtain the values of the unknowns by back-substitution. The method can be explained using the following example.

 

Example 2.2.1. Gauss Elimination and Back Substitution - A unique solution

Solve the linear system of four equations in three unknowns.

                                                  Formula

Solution.
The augmented matrix for the system is

                                             Formula

The Gauss elimination method is as follows.

Step 1 Locate the leftmost column that does not consist entirely of zeros.
          We locate the rst column.
Step 2 Interchange the top row with another row, if necessary, to bring a nonzero entry to the top of the column found in Step 1.
          Interchange the rst and the second row.

                                              Formula

Step 3 Add suitable multiples of the top row to the rows below so that all entries below the leading nonzero become zeros.
          Add 1 times the top row to the third row.
          Add -20 times the top row to the fourth row.

                                             Formula

Step 4 Now cover the top row in the matrix and begin again with Step 1 applied to the remaining row (Row 2 to 4). Continue in this way until the entire matrix is in the echelon or triangular form.
          Add -3 times the second row to the fourth row     .

                                        Formula

          Interchange the third row and the fourth row.

                                        Formula

The augmented matrix is now in the echelon form or triangular form. The corresponding system of equation is

                                             Formula

(We have discarded the last equation, Formula, since it will be satisfied automatically by the solutions of the remaining equations.)

The Gauss elimination has reduced the complex system of equation into a simpler one. The back-substitution method is then used to solve the simpler system.

     The back-substitution method is as follows.

Step 1 Solve the equation for the leading variables.

                                                  Formula

Step 2 Beginning with the bottom equation and working upward, successively substitute each equation into all the equations above it.
     Substitute x = 2 into the second equation yields

                                   Formula

          Substitute x = 4 and x = 2 into the first equation yields

                                             Formula

Step 3 Assign arbitrary values to the free variables, if any.
          Here, we do not have free variables.
We only have one solution, i.e. Formula.

 

Example 2.2.2. Gauss Elimination and Back Substitution - Infinitely many solutions

Solve the linear system of two equations in three unknowns.

                                                  Formula

Solution.
The augmented matrix for the system is

                                                       Formula

The Gauss elimination method reduces this matrix to an echelon form

                                                          Formula

The correspondings system of equations is

                                                       Formula

The back-substitution method is as follows.
Since

                                                       Formula

Then, Substitute Formula into the first equation yields

                    Formula

Assign Formula to Formula yields Formula.

Thus, the general solution is Formula, Formula, Formula.3

 

Exercise 2.2. Using the Gauss elimination and back substitution to solve the following system of equations

                                                  Formula

 

2.3 Existence of solutions based on echelon form

At the end of Gauss elimination (before the back substitution), the echelon matrix will have a corresponding reduced system in the form

                                                  Formula

where Formula.

 

There are three possible cases with respects to the solution of this system:

     (a)  No solution if Formula and one of Formula is not zero.

     (b)  Precisely one slution if Formula and Formula, if present, are zero.

     (c)  Infinitely many slutions if Formula and Formula, if present, are zero. Any of these solutions is obtained by assigned by assigning arbitrary values ti the free variables (See Example 2.2.2).

 

3 Gauss-Jordan Elimination Method

Gauss elimination method reduces the augmented matrix to an echelon form. Gauss-Jordan elimination method reduces the echelon form matrix further to a row-reduced echelon form, where the solution can be read out directly without the use of back-substitution. This form has the following properties

     (a)  If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leading 1.

     (b)  In any two successive rows, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row.

     (c)  Each column that contains a leading 1 has zeros everywhere else.

     (d)  If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix.

 

Example 3.1.

          Formula

where any real numbers substituted for the *'s.

For the first matrix, we can read the solution to the system of equations directly as Formula.

 

This method requires more effort to reduce the augmented matrix to the row-reduced echelon form. Thus, it is inferior to the Gauss elimination and back-substitution method because it requires more computational operations.

 

Example 3.2. Rework of Example 2.2.1 using Gauss-Jordan elimination

Solve the linear system of four equations in three unknowns.

                                        Formula

Solution.
The augmented matrix for the system is

                                   Formula

The echelon form of the matrix from the Gauss elimination method is

                                   Formula

The Gauss-Jordan elimination method reduces this matrix further V add extra steps to the Gauss elimination method in Example 2.2.1.

Step 5 Introduce a leading 1 to all the rows.
          Multiply the second row by 1/10.
          Multiply the third row by  -1/95.

                                             Formula

Step 6 Beginning with the last nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading 1's.
          Add -2.5 times the third row to the second row.

Formula

          Add -1 times the third row to the first row.
          Add 1 times the second row to the first row.

Formula

The matrix is in a reduced-row echelon form now and the solution can be read out as Formula.

 

Exercise 3. Using the Gauss-Jordan elimination to solve the system of equations in exercise 1.3.

 

 

 

4 Determinants

An n-th order determinant is a real number Formula associated with an Formula (square matrix) A.

It has an important application to

     (a) the theory of systems of linear equations,
     (b) in solving eigenvalue problems and differential equations,
     (c) in giving an explicit formula for the inverse of an invertible matrix.

 

4.1 Definitions of determinants

     First-order determinants

                                                  Formula

     Second-order determinants

                    Formula

     Third-order determinants

     Formula

The formula in (4) can be obtained by recopying the first and second columns as shown in Figure 1.
The determinant is then computed by summing the products on the rightward arrows and subtracting the products on the leftward arrows.

 Figure 1: Determinant of a 3 x 3 matrix     

 

Example 4.1. Evaluate the determinants of

                         Formula

Solution. Using (3) gives

               Formula

     Using the method of Figure 1 gives

                    Formula

Exercise 4.1. Find the determinant of

                                                            Formula

 

 

 

4.2 Cofactor Expansion - Minors and Cofactors

For a square matrix A

     1.  The  minor of entry Formula is denoted by Formula. It is defined to be the determinant of order Formula, namely the determinant of the submatrix obtained from A by deleting the j-th row and k-th column.

     2.  The  cofactor of entry Formula, denoted by Formula, is the number Formula, i.e. Formula . A quick way for determining whether to use the Formula or Formula is to use the fact that the sign relating Formula and Formula is in the j-th row and k-th column of the ``checkerboard" array

                                                  Formula

For example, Formula.

 

Exercise 4.2. Consider a matrix

                                                          Formula

Find Formula

 

 

 

 

 

 

 

 

 

4.3 Determinant of Any Order n

A determinant of order n is a scalar associated with an Formula matrix Formula, which is written

                                        Formula

The determinant can be computed using cofactor expansions, namely by multiplying the entries in any row (or column) by their cofactors and adding the resulting products; that is

Formula

where Formula is the cofactor of entry Formula.

 

Example 4.3. Let

                                             Formula

Evaluate Formula by cofactor expansion along the first row of A.

Solution. The cofactor of Formula is

               Formula

Similarly, the cofactor of Formula is

          Formula

Formula

Exercise 4.3. Consider the matrix A in exercise 4.1, using cofactor expansion to nd the determinant again.

 

 

 

 

 

 

 

 

 

 

4.4 General properties of determinants
4.4.1 Behaviour of an nth-order determinant under elementary row operations

     (a) Interchange of two rows multiplies the value of the determinant by  -1.

          Formula

 

     (b) Addition of a multiple of a row to another row does not alter the value of the determinant.

          Consider example 4.1, Formula. Adding the first and second row, i.e. Formula

               Formula

     (c) Multiplication of a row by a scalar c multiplies the value of the determinant by c.

                    Formula

 

4.4.2 Determinant of a triangular matrix

If A is an nxn triangular matrix (upper triangular, lower triangular or diagonal), then det(A) is the product of the entries on the main diagonal of the matrix; i.e.

                                                  Formula

 

4.4.3 Further properties of nth-order determinants

Suppose that A and B are n x n matrices and k is any scalar, then

     (a) Formula if Formula has a zero row or column.

     (b) Formula if Formula has proportional rows or columns.

     (c) Formula

     (d) Formula

     (e) In general, Formula

     (f) Formula

 

4.5 Evaluating determinants by row reduction

This method is an alternative to using cofactor expansion for evaluating determinants. It involves substantially less computation. It is well suited for computer evaluation of determinants because it is systematic and easily programmed.

 

The idea of this method is

     (a) to reduce the matrix to a triangular form by elementary row operations and then
     (b) to apply section 2.2 to compute the determinant of the triangular matrix and then
     (c) to apply section 2.1 to relate that determinant to that of the original matrix.
See below example.

 

Example 4.5. Evaluate det(A) where

                                             Formula

Solution. We will reduce A to an upper triangular form

Formula

 

Exercise 4.5. Consider the matrix in exercise 4.2, using the method in this section to nd the determinant.

 

 

 

 

 

 

 

 

 

 

5 Cramer's Rule

Cramers rule is a classical solution formula for linear systems. It is not practical in computations, but it is useful for studying the mathematical properties of a solution without the need for solving the system.

 

5.1 The Cramer's rule

If Formula is a system of Formula linear equations in Formula unknows such that Formula, then the system has a unique solution. This solution is

                                                  Formula

where Formula is the determinant obtained from Formula by replacing the k-th column in Formula by the column with the entries Formula.

 

Example 5.1. Use Cramer's rule to solve

                                                            Formula

Solution. Write the system of equations in matrix form,

Formula

Formula

 

Exercise 5.1. Solve the below system of equations by Cramer's rule.

                                                  Formula

 

 

 

 

Comments (0)

You don't have permission to comment on this page.