Linear Algebra (continued)

**1 Systems of Linear Equations**

**1.1 General Form**

A linear equation in the* n* variables or unknows is an equation of the form

where and are real constants.

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

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 has the form

where and are constans.

**Note.**

(a) The are called the of the system.

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

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

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

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

The system has a solution , , since these values satisfy both equations. However, , , 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 .

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

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

where the **coefficient matrix A** is the matrix

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

The matrix

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

is

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

**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.

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.**

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.

**Solution.**

The augmented matrix for the system is

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.*

**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.*

* *

**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 .*

* *

* Interchange the third row and the fourth row.*

* *

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

(We have discarded the last equation, , 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.

**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*

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

**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. .

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

Solve the linear system of two equations in three unknowns.

**Solution.**

The augmented matrix for the system is

The Gauss elimination method reduces this matrix to an echelon form

The correspondings system of equations is

The back-substitution method is as follows.

Since

Then, Substitute into the first equation yields

Assign to yields .

Thus, the general solution is , , .3

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

**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

where .

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

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

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

(c) Infinitely many slutions if and , 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.**

where any real numbers substituted for the *'s.

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

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.

**Solution.**

The augmented matrix for the system is

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

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.*

**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.*

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

* Add 1 times the second row to the first row.*

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

**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 associated with an (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

Second-order determinants

Third-order determinants

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

**Solution**. Using (3) gives

Using the method of Figure 1 gives

**Exercise 4.1**. Find the determinant of

**4.2 Cofactor Expansion - Minors and Cofactors**

For a square matrix A

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

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

For example, .

**Exercise 4.2**. Consider a matrix

Find

**4.3 Determinant of Any Order ***n*

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

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

where is the cofactor of entry .

**Example 4.3**. Let

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

Solution. The cofactor of is

Similarly, the cofactor of is

**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 ***n*th-order determinant under elementary row operations

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

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

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

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

**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.

**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) if has a zero row or column.

(b) if has proportional rows or columns.

(c)

(d)

(e) In general,

(f)

**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

**Solution**. We will reduce A to an upper triangular form

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

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

**Example 5.1**. Use Cramer's rule to solve

**Solution**. Write the system of equations in matrix form,

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

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