MATH 250 Linear Equations and Matrices

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MATH 250Linear Equations and Matrices
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Topics

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

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Systems of Linear Equations

• System of m equations in n unknowns

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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

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Matrices

• Systems of Equations
• Since the solution of the system involves the a
  and b values only, will often work with the
  augmented matrix

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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

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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

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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).
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Example
Use matrices to solve the system
Solution
Step 1 Write the augmented matrix for the system.
Linear System
Augmented Matrix
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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.)
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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.

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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.
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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.

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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
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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.
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The corresponding system of equation is
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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.
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