MATH 250Linear Equations and Matrices

Topics

- Preliminaries
- Systems of Linear Equations
- Matrices
- Algebraic Properties of Matrix Operations
- Special Types of Matrices and Partitioned

Matrices - Matrix Transformations

Systems of Linear Equations

- System of m equations in n unknowns

Systems of Linear Equations

- Comments
- If a system has a solution, call it consistent
- If a system doesnt have a solution, call it

inconsistent - If , the system is

called homogeneous. A homogeneous system always

has the trivial solution - If two systems have the same solution, then they

are called equivalent. The solution strategy for

linear systems is to transform the system through

a series of equivalent systems until the solution

is obvious

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Matrices

- Systems of Equations
- Consider
- Define
- Express system as AX B

Matrices

- Systems of Equations
- Since the solution of the system involves the a

and b values only, will often work with the

augmented matrix

Systems of Linear Equations

- Elementary Operations on Systems
- Switch two equations
- Multiply an equation by nonzero constant
- Add multiple of one equation to another
- The application of any combination of elementary
- operations to a linear system yields a new linear

system - that is equivalent to the first

Inverting a Matrix

- Usually not a good idea to compute xA-1b
- Inefficient
- Prone to roundoff error
- In fact, compute inverse using linear solver
- Solve Axibi where bi are columns of identity,xi

are columns of inverse - Many solvers can solve several R.H.S. at once

Solving Linear Systems Using Gaussian Elimination

- Write the augmented matrix for the system
- Use matrix row operations to simplify the matrix

to one with 1s down the diagonal from upper left

to lower right, and 0s below the 1s - Write the system of linear equations

corresponding to the matrix in step 2, and use

back-substitution to find the systems solutions

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Theorem 1 Equivalent systems and equivalent

matrices

- If the augmented coefficient matrices of two

linear systems are row equivalent, then the two

systems have the same solution set.

Definition Echelon Matrix

The matrix E is called an echelon matrix provided

it has the following two properties 1. Every

row of E that consists entirely of zeros (if any)

lies beneath every row that contains a nonzero

element. 2. In each row of E that contains a

nonzero element, the nonzero element lies

strictly to the right of the first nonzero

element in the preceding row (if there is a

preceding row).

Example

Use matrices to solve the system

Solution

Step 1 Write the augmented matrix for the system.

Linear System

Augmented Matrix

Example cont.

Solution

Step 2 Use matrix row operations to simplify the

matrix to one with 1s down the diagonal from

upper left to lower right, and 0s below the 1s.

Our goal is to obtain a matrix of the form

.

Our first step in achieving this goal is to get 1

in the top position of the first column.

To get 1 in this position, we interchange rows 1

and 2. (We could also interchange rows 1 and 3 to

attain our goal.)

Example cont.

Solution

Now we want to get 0s below the 1 in the first

column.

Lets first get a 0 where there is now a 3. If we

multiply the top row of numbers by 3 and add

these products to the second row of numbers, we

will get 0 in this position. The top row of

numbers multiplied by 3 gives -3(1) or 3, -3(1)

or 3, -3(2) or 6, -3(19) or 57.

Now add these products to the corresponding

numbers in row 2. Notice that although we use row

1 to find the products, row 1 does not change.

Example cont.

Solution

We are not yet done with the first column. If we

multiply the top row of numbers by 1 and add

these products to the third row of numbers, we

will get 0 in this position. The top row of

numbers multiplied by 1 gives -1(1) or

1, -1(1) or 1, -1(2) or 2, -1(19) or 19.

Now add these products to the corresponding

numbers in row 3.

We move on to the second column. We want 1 in the

second row, second column.

Example cont.

Solution

To get 1 in the desired position, we multiply 2

by its reciprocal, -1/2. Therefore, we multiply

all the numbers in the second row by 1/2 to get

We are not yet done with the second column. If we

multiply the top row of numbers by 2 and add

these products to the third row of numbers, we

will get 0 in this position. The second row of

numbers multiplied by 2 gives -2(0) or 0, -2(1)

or 2, -2(2) or 4, -2(13) or 26.

Now add these products to the corresponding

numbers in row 3.

Example cont.

Solution

We move on to the third column. We want 1 in the

third row, third column.

To get 1 in the desired position, we multiply 4

by its reciprocal, -1/4. Therefore, we multiply

all the numbers in the third row by 1/4 to get

We now have the desired matrix with 1s down the

diagonal and 0s below the 1s.

Step 3 Write the system of linear equations

corresponding to the matrix in step 2, and use

back-substitution to find the systems solution.

The system represented by the matrix in step 2 is

Example cont.

Solution

We immediately see that the value for z is 5. To

find y, we back-substitute 5 for z in the second

equation.

y 2z 13 Equation 2 y 2(5) 13 Substitute

5 for x. y 3 Solve for y.

Finally, back-substitute 3 for y and 5 for z in

the first equation

x y 2z 19 Equation 1 x 3 2(5)

19 Substitute 3 for y and 5 for x. x 13 19

Multiply and add. x 6 Subtract 13 from both

sides.

The solution set for the original system is (6,

3, 5).

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Gaussian Elimination

- Definition
- A matrix is in echelon form if
- Any rows consisting entirely of zeros are grouped

at the bottom of the matrix. - The first nonzero element of each row is 1. This

element is called a leading 1. - The leading 1 of each row after the first is

positioned to the right of the leading 1 of the

previous row. - (This implies that all the elements below a

leading 1 are zero.)

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Example 1

Solving the following system of linear equations

using the method of Gaussian elimination.

Solution

We have arrived at the echelon form.

The corresponding system of equation is

Example 2

Solving the following system of linear equations

using the method of Gaussian elimination,

performing back substitution using matrices.

Solution

This marks the end of the forward elimination of

variables from equations. We now commence the

back substitution using matrices.

2