LinearEqn n Matrices

7/29/2019 LinearEqn n Matrices

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5

Systems of Linear Equations: An Introduction

Unique Solutions

Underdetermined and

Overdetermined Systems

Matrices

Multiplication of Matrices

The Inverse of a Square Matrix

Systems of Linear Equations and Matrices


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5.1Systems of Linear Equations:

An Introduction


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Systems of Equations

Recall that a system of two linear equations in twovariables may be written in the general form

where a, b, c, d, h, and kare real numbers and neithera and b nor c and dare both zero.

Recall that the graph of each equation in the system is a

straight line in the plane, so that geometrically, the

solution to the system is the point(s) of intersection of thetwo straight linesL1 andL2, represented by the first and

second equations of the system.

ax by h

cx dy k


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Given the two straight linesL1 andL2, one and only one ofthe following may occur:

1.L1 andL2 intersect at exactly one point.

y

x

L1

L2

Unique

solution(x1,y1)

(x1,y1)

x1

y1


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2.L1 andL2 are coincident.

y

x

L1,L2

Infinitely

manysolutions


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3.L1 andL2 are parallel.

y

x

L1L2

No

solution


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Example:

A System of Equations With Exactly One Solution

Consider the system

Solving the first equation fory in terms ofx, we obtain

Substituting this expression fory into the second equation

yields

2 1

3 2 12

x y

x y

2 1y x

3 2(2 1) 12

3 4 2 12

7 14

2

x x

x x

x

x


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Example:

A System of Equations With Exactly One Solution

Finally, substituting this value ofx into the expression for

y obtained earlier gives

Therefore, the unique solution of the system is given by

x = 2 andy = 3.

2 1

2(2) 1

3

y x


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Example:

A System of Equations With Infinitely Many Solutions

Consider the system


Substituting this expression fory into the second equationyields

which is a true statement.

This result follows from the fact that the second equation

is equivalent to the first.

2 16 3 3

x yx y

2 1y x

6 3(2 1) 3

6 6 3 3

0 0

x x

x x


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Example:


Thus, any order pair of numbers (x,y) satisfying theequation y =2x1 constitutes a solution to the system.

By assigning the valuettox, where tis any real number,

we find that y =2t1 and so the ordered pair (t, 2t1)

is a solution to the system. The variable tis called a parameter.

For example:

Setting t= 0, gives the point (0,1) as asolution of the

system.

Setting t= 1, gives the point (1, 1) as another solution of

the system.


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6

5

4

3

2

1

11 2 3 4 5 6

Example:


Since trepresents any real number, there are infinitely

many solutions of the system.

Geometrically, the two equations in the system representthe same line, and all solutions of the system are pointslying on the line:

y

x

2 1

6 3 3

x y

x y


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Example:

A System of Equations That Has No Solution

Consider the system


Substituting this expression fory into the second equationyields

which is clearlyimpossible.

Thus, there is no solution to the system of equations.

2 16 3 12x yx y

2 1y x

6 3(2 1) 12

6 6 3 12

0 9

x x

x x


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1 2 3 4 5 6

Example:


To interpret the situation geometrically, cast both

equations in the slope-intercept form, obtaining

y = 2x1 and y = 2x4

which shows that the lines are parallel.

Graphically:

6

5

4

3

2

1

1

y

x

2 1x y

6 3 12x y


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Unique Solutions

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1


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The Gauss-Jordan Method

The Gauss-Jordan elimination method is a technique forsolving systems of linear equations of any size.

The operations of the Gauss-Jordan method are

1. Interchange any two equations.

2. Replace an equation by a nonzero constant multiple of

itself.

3. Replace an equation by the sum of that equation and aconstant multiple of any other equation.


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Example

Solve the following system of equations:

Solution

First, we transform this system into an equivalent systemin which the coefficient ofx in the first equation is 1:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

Multiply the

equation by 1/2Toggle slidesback and forth tocompare before

and changes


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Example


Solution

Next, we eliminate the variablex from all equations exceptthe first:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 113 8 5 27

2 2

x y zx y z

x y z

Replace by the sum of

3 X the first equation

+ the second equation

Toggle slidesback and forth tocompare before

and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 112 4 6

2 2

x y zy z

x y z


3 the first equation

+ the second equation


and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 112 4 6

2 2

x y zy z

x y z

Replace by the sumof the first equation

+ the third equation


and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Replace by the sumof the first equation

+ the third equation


and changes


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Example


Solution

Then we transform so that the coefficient ofy in thesecond equation is 1:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Multiply the second

equation by 1/2


and changes


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Example


Solution

Then we transform so that the coefficient ofy in thesecond equation is 1:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z

Multiply the second

equation by 1/2


and changes


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Example


Solution

We now eliminatey from all equations except the second:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z


the first equation +

(2) the second equation


and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

7 17

2 3

3 5 13

x z

y z

y z


the first equation +



and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

7 17

2 3

3 5 13

x z

y z

y z

Replace by the sum ofthe third equation +



and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

7 17

2 3

11 22

x z

y z

z

Replace by the sum ofthe third equation +



and changes


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Example


Solution

Now we transform so that the coefficient ofzin the thirdequation is 1:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

7 17

2 3

11 22

x z

y z

z

Multiply the thirdequation by 1/11


and changes


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Example


Solution

Now we transform so that the coefficient ofzin the thirdequation is 1:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

7 17

2 3

2

x z

y z

z

Multiply the thirdequation by 1/11


and changes


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Example


Solution

We now eliminatezfrom all equations except the third:

2 4 6 223 8 5 27

2 2

x y zx y z

x y z


the first equation +(7) the third equation

7 17

2 3

2

x z

y z

z


and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

3

2 3

2

x

y z

z


the second equation +

2the third equation


and changes


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Example


Solution


2 4 6 223 8 5 27

2 2

x y zx y z

x y z

3

1

2

x

y

z


the second equation +

2the third equation


and changes


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Example


Solution

Thus, the solution to the system isx = 3,y = 1, andz= 2.

2 4 6 223 8 5 27

2 2

x y zx y z

x y z

3

1

2

x

y

z


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Augmented Matrices

Matrices are rectangular arrays of numbers that can aid

us by eliminating the need to write the variables at eachstep of the reduction.

For example, the system

may be represented by the augmentedmatrixCoefficient

Matrix

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

1 1 2 2


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Matrices and Gauss-Jordan

Every step in the Gauss-Jordan elimination method can be

expressed with matrices, rather than systems of equations,

thus simplifying the whole process:

Steps expressed as systems of equations:

Steps expressed as augmented matrices:

1 2 3 11

3 8 5 27

1 1 2 2

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z


and changes


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1 2 3 11

0 2 4 6

1 1 2 2

2 3 11

2 4 6

2 2

x y z

y z

x y z


and changes


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1 2 3 11

0 2 4 6

0 3 5 13

2 3 11

2 4 6

3 5 13

x y z

y z

y z


and changes


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1 2 3 11

0 1 2 3

0 3 5 13

2 3 11

2 3

3 5 13

x y z

y z

y z


and changes


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1 0 7 17

0 1 2 3

0 3 5 13

7 17

2 3

3 5 13

x z

y z

y z


and changes


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1 0 7 17

0 1 2 3

0 0 11 22

7 17

2 3

11 22

x z

y z

z


and changes

i G


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1 0 7 17

0 1 2 3

0 0 1 2

7 17

2 3

2

x z

y z

z


and changes

M i d G J d


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1 0 0 3

0 1 2 3

0 0 1 2

3

2 3

2

x

y z

z


and changes

M i d G J d


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1 0 0 3

0 1 0 1

0 0 1 2

3

1

2

x

y

z

Row Reduced Form

of the Matrix


and changes


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Row-Reduced Form of a Matrix

Each row consisting entirely ofzeros lies below all

rows having nonzero entries.

The firstnonzero entry in each nonzero row is 1

(called a leading1).

In any two successive (nonzero) rows, the leading1

in the lower row lies to the right of the leading1 in

the upper row.

If a column contains a leading1, then the otherentries in that column are zeros.


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Row Operations

1. Interchange any two rows.

2. Replace any row by a nonzero constant

multiple of itself.

3. Replace any row by the sum of that rowand a constant multiple of any other row.


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Terminology for the

Gauss-Jordan Elimination Method

Unit Column

A column in a coefficient matrix is in unit form

ifone of the entries in the column is a 1 and the

other entries are zeros.

Pivoting

The sequence ofrow operations that transforms

a given column in an augmented matrix into a

unit column.


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Notation for Row Operations

LettingRi denote the ithrow of a matrix, we write

Operation 1: Ri Rjto mean:

Interchange row i with rowj.

Operation 2: cRito mean:replace row i with ctimes row i.

Operation 3: Ri + aRj to mean:

Replace row i with the sum of row i

and atimes rowj.


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Example

Pivot the matrix about the circled element

Solution

3 5 9

2 3 5

3 5 9

2 3 5

113

R 53

31

52 3

2 12R R 53

13

1 3

0 1


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The Gauss-Jordan Elimination Method

1. Write the augmented matrix corresponding tothe linear system.

2. Interchange rows, if necessary, to obtain an

augmented matrix in which the first entry in

the first row is nonzero. Then pivot the matrixabout this entry.

3. Interchange thesecond row with any row below

it, if necessary, to obtain an augmented matrix

in which the second entry in the second row is

nonzero. Pivot the matrix about this entry.

4. Continue until the final matrix is in row-

reduced form.


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Example

Use the Gauss-Jordan elimination method to solve the

system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 1 3

1 2 3 8

1 2R R

Toggle slides back andforth to compare before

and after matrix changes


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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

2 2 1 3

1 2 3 8

2 12R R

3 1R R

1 2R R




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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

2 3R R1

22R

1 0 9 12

0 2 12 4

0 2 19 27




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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

122

R

1 0 9 12

0 1 6 2

0 2 19 27

3 2R R




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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 31 31

3 2R R1

331R




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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 1 1

1331

R1 3

9R R

2 36R R



E l


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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

2 36R R

1 39R R



E l


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Example


system of equations

Solution

The solution to the system is thusx = 3,y = 4, andz= 1.

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1


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Underdetermined and Overdetermined systems

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

3 1 2 1

2 3 5 3

1

x z

y z

0

1

x z

y z

A System of Equations


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with an Infinite Number of Solutions

Solve the system of equations given by

Solution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

3 1 2 1

2 3 5 3

2 13R R

3 12R R





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Solution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

0 7 7 7

0 1 1 1

2 13R R

3 12R R

127

R





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Solution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 2 3 2

0 1 1 1

0 1 1 1

127

R1 22R R

3 2R R





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Solution

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0

1 22R R

3 2R R





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Solution

Observe that row three reads 0 = 0, which is true but

of no use to us.

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0



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Solution

This last augmented matrix is in row-reduced form.

Interpreting it as a system of equations gives a system oftwo equations in three variablesx,y, andz:

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1 0 1 0

0 1 1 1

0 0 0 0

0

1

x z

y z



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Solution

Lets single out a single variablesay,zand solve forxandy in terms of it.

If we assign a particular value ofzsay,z= 0we obtain

x = 0 andy =

1, giving the solution(0,

1, 0).

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

0

1

x z

y z

(0) 0

(0) 1 1



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Solution


If we instead assignz= 1, we obtain the solution(1, 0, 1).

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

0

1

x z

y z

(1) 1

(1) 1 0



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Solution


In general, we setz= t, where trepresents any real number

(called the parameter) to obtain the solution(t, t

1, t).

2 3 2

3 2 1

2 3 5 3

x y z

x y z

x y z

1

x z

y z

0

1

x z

y z

( )

( ) 1 1

t t

t t

A S f E i Th H N S l i


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Solution

1

3 4

5 5 1

x y z

x y z

x y z

1 1 1 1

3 1 1 4

1 5 5 1

2 13R R

3 1R R



A S f E i Th H N S l i


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Solution

1 1 1 1

0 4 4 1

0 4 4 2

2 13R R

3 1R R

3 2R R

1

3 4

5 5 1

x y z

x y z

x y z



A S t f E ti Th t H N S l ti


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Solution

1 1 1 1

0 4 4 1

0 0 0 1

3 2R R

1

3 4

5 5 1

x y z

x y z

x y z



A S t f E ti Th t H N S l ti


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Solution

Observe that row three reads 0x + 0y + 0z=1 or 0 =1!

We therefore conclude the system is inconsistent and has

no solution.

1 1 1 1

0 4 4 1

0 0 0 1

1

3 4

5 5 1

x y z

x y z

x y z


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Systems with no Solution

If there is a row in the augmented matrix

containing all zeros to the left of the vertical line

and a nonzero entry to the right of the line, then

the system of equations has no solution.


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Theorem 1

a. If the number of equations is greater than or

equal to the number of variables in a linear

system, then one of the following is true:

i. The system has no solution.ii. The system has exactly one solution.

iii. The system has infinitely many solutions.

b. If there are fewerequations than variables in

a linear system, then the system either has nosolution or it has infinitely many solutions.


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5.4

Matrices

2 3

2 3

X B A

X A B

2 3

2 3

X B A

X A B

3 4 3 231 2 1 2

3 4 3 231 2 1 2

9 12 3 2

3 6 1 2

9 12 3 2

3 6 1 2

6 10

2 4

6 10

2 4

6 101

2 42X

6 101

2 42X

3 5

1 2

3 5

1 2

M i


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Matrix

A matrix is an ordered rectangular array of numbers.

A matrix with mrows and ncolumns has size m n.

The entry in the ithrow andjthcolumn is denoted by aij.

Applied Example: Organizing Production Data


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Applied Example:Organizing Production Data

The Acrosonic Company manufactures four differentloudspeaker systems at three separate locations.

The companys May output is as follows:

If we agree to preserve the relative location of each entryin the table, we can summarize the set of data as follows:

Model A Model B Model C Model D

Location I 320 280 460 280

Location II 480 360 580 0Location III 540 420 200 880

320 280 460 280

480 360 580 0

540 420 200 880

Applied Example: Organizing Production Data


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We have Acrosonics May output expressed as a matrix:

a. What is the size of the matrixP?

Solution

MatrixPhas three rows and four columns and hence

has size3 4.

320 280 460 280

480 360 580 0

540 420 200 880

P



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pp ed a p e O g g oduc o


b. Find a24 (the entry in row 2 and column 4 of thematrixP) and give an interpretation of this number.

Solution

The required entry lies in row 2 and column 4, and is

the number 0. This means that no model Dloudspeaker system was manufactured at location IIin May.

320 280 460 280

480 360 580 0

540 420 200 880

P



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pp p g g


c. Find the sum of the entries that make up row1 ofPand interpret the result.

Solution

The required sum is given by

320 + 280 + 460 + 280 = 1340

which gives the total number of loudspeaker systemsmanufactured at location I in May as 1340 units.

320 280 460 280

480 360 580 0

540 420 200 880

P



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pp p g g


d. Find the sum of the entries that make up column4 ofPand interpret the result.

Solution

The required sum is given by

280 + 0 + 880 = 1160

giving the output ofModel D loudspeaker systems atalllocations in May as 1160 units.

320 280 460 280

480 360 580 0

540 420 200 880

P

E lit f M t i


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Equality of Matrices

Two matrices are equal if they have the same size

and their corresponding entries are equal.

Example


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Example

Solve the following matrix equation forx,y, andz:

Solution

Since the corresponding elements of the two matrices mustbe equal, we find thatx = 4,z= 3, andy1 = 1, ory = 2.

1 3 1 4

2 1 2 2 1 2

x z

y

Addition and Subtraction of Matrices


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Addition and Subtraction of Matrices

IfA andB are two matrices of the same size, then:

1. The sumA + B is the matrix obtained by adding

the corresponding entries in the two matrices.

2. The differenceA

B is the matrix obtained by

subtracting the corresponding entries inB from

those inA.



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g g

The total output of Acrosonic for May is

The total output of Acrosonic for June is

Find the total output of the company for May and June.


Location I 210 180 330 180

Location II 400 300 450 40

Location III 420 280 180 740


Location I 320 280 460 280

Location II 480 360 580 0

Location III 540 420 200 880



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Solution

Expressing the output for May and June as matrices: The total output of Acrosonic for May is

The total output of Acrosonic for June is

320 280 460 280

480 360 580 0

540 420 200 880

A

210 180 330 180400 300 450 40

420 280 180 740

B



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Solution

The total output of the company for May and June isgiven by the matrix

320 280 460 280 210 180 330 180

480 360 580 0 400 300 450 40

540 420 200 880 420 280 180 740

530 460 790 460

880 660 1030 40

960 700 380 1620

A B

Laws for Matrix Addition


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Laws for Matrix Addition

IfA,B, and Care matrices of the same size, then

1. A +B = B + A Commutative law

2. (A +B) + C= A + (B + C) Associative law

Transpose of a Matrix


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Transpose of a Matrix

IfA is an m n matrix with elements aij,

then the transpose ofA is the n m matrix

ATwith elements aji.

Example


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Find the transpose of the matrix

Solution

The transpose of the matrixA is

1 2 3

4 5 6

7 8 9

A

1 4 7

2 5 8

3 6 9

TA

S l P d t


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Scalar Product

IfA is a matrix and c is a real number, then

the scalar productcA is the matrix obtained

by multiplyingeach entry ofA by c.

Example


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Given

find the matrixXthat satisfies 2X+ B = 3A

Solution

2 3

2 3

X B A

X A B

3 4 3 23

1 2 1 2

9 12 3 2

3 6 1 2

6 10

2 4

6 101

2 42X

3 5

1 2

3 4 3 2

1 2 1 2A B

and

Applied Example:Production Planning


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The management of Acrosonic has decided to increase itsJuly production ofloudspeaker systems by 10%

(over June output). Find a matrix giving the targeted production for July.

Solution

We have seen that Acrosonics total output for June may

be represented by the matrix

210 180 330 180

400 300 450 40

420 280 180 740

B

Applied Example:Production Planning


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The management of Acrosonic has decided to increase itsJuly production ofloudspeaker systems by 10%

(over June output). Find a matrix giving the targeted production for July.

Solution

The required matrix is given by

210 180 330 180

(1.1) 1.1 400 300 450 40

420 280 180 740

B

231 198 363 198

440 330 495 44

462 308 198 814


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5.5Multiplication of Matrices

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a aA B b b b b

a a ab b b b

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a aA B b b b b

a a ab b b b

Size ofA (2 3) (3 4) Size ofB

(2 4)

Size ofAB

Same

Multiplying a Row Matrix by a Column Matrix


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If we have a row matrix of size 1n,

And a column matrix of size n 1,

Then we may define the matrix product ofA and B, written

AB, by

1 2 3[ ]nA a a a a

1

2

3

n

bb

B b

b

1

2

1 2 3 3 1 1 2 2 3 3[ ]n n n

n

b

b

AB a a a a b a b a b a b a b

b

Example


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Let

Find the matrix productAB.

Solution

2

3

[1 2 3 5] 0

1

a d nA B

2

3[1 2 3 5] (1)(2) ( 2)(3) (3)(0) (5)( 1) 9

01

AB

Dimensions Requirement


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for Matrices Being Multiplied

Note from the last example that for the multiplication to

be feasible, the number of columns of the row matrixA

must be equalto the number of rows of the column

matrixB.

Dimensions of the Product Matrix


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From last example, note that the product matrixAB has

size 1 1. This has to do with the fact that we are multiplying a row

matrix with a column matrix.

We can establish the dimensionsof a product matrix

schematically:

Size ofA (1

n) (n

1) Size ofB

Size ofAB

(1 1)

Same



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More generally, ifA is a matrix of size mn andB is amatrix of size np, then the matrix product ofA andB,

AB, is defined and is a matrix ofsizemp.

Schematically:

The number of columns ofA must be the sameas the

number of rows ofB for the multiplication to be feasible.

Size ofA (mn) (np) Size ofB

Size ofAB(mp)

Same

Mechanics of Matrix Multiplication


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To see how to compute the product of a 2 3 matrix A anda 3 4 matrixB, suppose

From the schematic

we see that the matrix product C = AB is feasible (since thenumber ofcolumns ofAequals the number ofrows ofB) andhas size2 4.

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

Size ofA (2 3) (3 4) Size ofB

Size ofAB

(2 4)

Same



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Thus,

To see how to calculate the entries ofCconsider entryc11

:

11 12 13 14

21 22 23 24

c c c cC

c c c c

11

11 11 12 13 21 11 11 12 21 13 31

31

[ ]

b

c a a a b a b a b a b

b

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



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Thus,

Now consider calculating the entryc12

:

11 12 13 14

21 22 23 24

c c c cC

c c c c

12

12 11 12 13 22 11 12 12 22 13 32

32

[ ]

b


b

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



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Thus,

Now consider calculating the entryc21

:

11 12 13 14

21 22 23 24

c c c cC

c c c c

11

21 21 22 23 21 21 11 22 21 23 31

31

[ ]

b


b

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b



108/145


Thus,

Other entries are computed in a similar manner.

11 12 13 14

21 22 23 24

c c c cC

c c c c

11 12 13 14

11 12 13

21 22 23 24

21 22 23

31 32 33 34

b b b ba a a

A B b b b ba a a

b b b b

Example


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Let

ComputeAB.

Solution

Since the number of columns ofA is equal to the numberof rows ofB, the matrix product C = ABis defined.

The size ofCis 2 3.

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

Example


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Let

ComputeAB.

Solution

Thus,

Calculate all entries for C:

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

11 12 13

21 22 23

1 3 33 1 44 1 2

1 2 32 4 1

c c cC AB

c c c

11

1

[3 1 4] 4 (3)(1) (1)(4) (4)(2) 15

2

c

Example


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Let

ComputeAB.

Solution

Thus,


1 3 33 1 4

4 1 21 2 3

2 4 1

A B

12 13

21 22 23

1 3 3 153 1 44 1 2

1 2 32 4 1

c cC AB

c c c

12

3

[3 1 4] 1 (3)(3) (1)( 1) (4)(4) 24

4

c

Example


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Let

ComputeAB.

Solution

Thus,


1 3 33 1 4

4 1 21 2 3

2 4 1

A B

13

21 22 23

1 3 3 15 243 1 44 1 2

1 2 32 4 1

cC AB

c c c

13

3

[3 1 4] 2 (3)( 3) (1)(2) (4)(1) 3

1

c

Example


113/145

Let

ComputeAB.

Solution

Thus,


1 3 33 1 4

4 1 21 2 3

2 4 1

A B

21 22 23

1 3 3 15 24 33 1 44 1 2

1 2 32 4 1

C ABc c c

21

1

[ 1 2 3] 4 ( 1)(1) (2)(4) (3)(2) 13

2

c

Example


114/145

Let

ComputeAB.

Solution

Thus,


1 3 33 1 4

4 1 21 2 3

2 4 1

A B

22 23

1 3 3 15 24 33 1 44 1 2

131 2 32 4 1

C ABc c

22

3

[ 1 2 3] 1 ( 1)(3) (2)( 1) (3)(4) 7

4

c

Example


115/145

Let

ComputeAB.

Solution

Thus,


1 3 33 1 4

4 1 21 2 3

2 4 1

A B

23

1 3 3 15 24 33 1 44 1 2

13 71 2 32 4 1

C ABc

23

3

[ 1 2 3] 2 ( 1)( 3) (2)(2) (3)(1) 10

1

c

Example


116/145

Let

ComputeAB.

Solution

Thus,

1 3 33 1 4

4 1 21 2 3

2 4 1

A B

1 3 33 1 4 15 24 34 1 2

1 2 3 13 7 102 4 1

C AB

Laws for Matrix Multiplication


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If the products and sums are defined for the

matricesA,B, and C, then

1. (AB)C=A(BC) Associative law

2. A(B + C) =AB +AC Distributive law

Identity Matrix


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The identity matrix ofsizen is given by

nrows

ncolumns

1 0 0

0 1 0

0 0 1

nI

Properties of the Identity Matrix


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The identity matrix has the properties that

InA = A for any nrmatrixA.

BIn= B for anys n matrixB. In particular, ifA is a square matrix of

size n, then

n nI A AI A

Example

Let


120/145

Let

Then

So, I3A =AI3=A.

1 3 1

4 3 21 0 1

A

3

1 0 0 1 3 1 1 3 1

0 1 0 4 3 2 4 3 2

0 0 1 1 0 1 1 0 1

I A A

3

1 3 1 1 0 0 1 3 1

4 3 2 0 1 0 4 3 2

1 0 1 0 0 1 1 0 1

AI A

Matrix Representation


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A system of linear equations can be expressed in the formofan equation of matrices. Consider the system

The coefficients on the left-hand side of the equation canbe expressed as matrixA below, the variables as matrixX,and the constants on right-hand side of the equation asmatrixB:

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 1 6

3 6 5 1

1 3 7 0

x

A X y B

z



122/145


The matrix representation of the system of linearequations is given byAX= B, or

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 1 6

3 6 5 1

1 3 7 0

x

y

z



123/145


To confirm this, we can multiply the two matrices on theleft-hand side of the equation, obtaining

which, by matrix equality, is easily seen to be equivalent tothe given system of linear equations.

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

2 4 6

3 6 5 1

3 7 0

x y z

x y z

x y z

5 6


124/145

5.6The Inverse of a Square Matrix

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

x

y

z

x

y

z

Inverse of a Matrix


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LetA be a square matrix of size n.

A square matrixA1of size n such that

is called the inverse ofA.

Not every matrix has an inverse.

A square matrix that has an inverse is

said to be nonsingular.

A square matrix that does not have an

inverse is said to be singular.

1 1

nA A AA I

Example:A Nonsingular Matrix


126/145

The matrix has a matrix

as its inverse.

This can be demonstrated by multiplying them:

1 2

3 4A

1

3 12 2

2 1A

1

3 12 2

2 11 2 1 0

3 4 0 1AA I

1

3 12 2

2 1 1 2 1 0

3 4 0 1A A I

Example:A Singular Matrix


127/145

The matrix does not have an inverse.

IfB had an inverse given by where

a, b, c, and dare some appropriate numbers, then bydefinition of an inverse we would haveBB1 = I.

That is

implying that 0 = 1, which is impossible!

0 1

0 0B

0 1 1 0

0 0 0 1

1 0

0 0 0 1

a b

c d

c d

1a b

Bc d

Finding the Inverse of a Square Matrix


128/145

Given the n n matrixA:

1. Adjoin the n n identity matrixIto obtain

the augmented matrix[A|I].

2. Use a sequence ofrow operations to reduce[A|I] to the form [I|B] if possible.

Then the matrixB is the inverse ofA.

Example


129/145

Find the inverse of the matrix

Solution

We form the augmented matrix

2 1 1

3 2 12 1 2

A

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

Example


130/145


Solution

And use the Gauss-Jordan elimination method to reduce it

to the form [I|B]:

2 1 1

3 2 12 1 2

A

2 1 1 1 0 0

3 2 1 0 1 0

2 1 2 0 0 1

1 2R R


and changes

Example


131/145


Solution


to the form [I|B]:

2 1 1

3 2 12 1 2

A

1 1 0 1 1 0

3 2 1 0 1 0

2 1 2 0 0 1

1 2R R

1

2 3

3 1

3

2

R

R R

R R

Toggle slidesback and forth to

compare beforeand changes

Example


132/145


Solution


to the form [I|B]:

2 1 1

3 2 12 1 2

A

1 1 0 1 1 0

0 1 1 3 2 0

0 1 2 2 2 1

1

2 3

3 1

3

2

R

R R

R R

1 2

2

3 2

R R

R

R R

Toggle slidesback and forth to

compare beforeand changes

Example


133/145

1 2

2

3 2

R R

R

R R


Solution


to the form [I|B]:

2 1 1

3 2 12 1 2

A

1 0 1 2 1 0

0 1 1 3 2 0

0 0 1 1 0 1

1 3

2 3

R R

R R


and changes

Example


134/145


Solution


to the form [I|B]:

2 1 1

3 2 12 1 2

A

1 0 0 3 1 1

0 1 0 4 2 1

0 0 1 1 0 1

1 3

2 3

R R

R R

In B


and changes

Example


135/145


Solution

Thus, the inverse ofA is the matrix

2 1 1

3 2 12 1 2

A

3 1 1

4 2 1

1 0 1

1 A

A Formula for the Inverse of a 2 2 Matrix


136/145

Let

Suppose D = adbc is not equal to zero.

ThenA1exists and is given by

a bA

c d

1 1 d bAc aD

Example


137/145

Find the inverse of

Solution

We first identifya, b, c, and das being 1, 2, 3, and 4

respectively.

We then compute

D = adbc= (1)(4)(2)(3) = 46 =2

1 2

3 4A

Example


138/145

Find the inverse of

Solution

Next, we substitute the values1, 2, 3, and 4 instead ofa, b, c, and d, respectively, in the formula matrix

to obtainthe matrix

1 2

3 4A

4 2

3 1

d b

c a

Example


139/145

Find the inverse of

Solution

Finally, multiplying this matrix by 1/D, we obtain

1 2

3 4A

1

3 12 2

2 14 21 1

3 12

d bA

c aD

Using Inverses to Solve Systems of Equations


140/145

IfAX=B is a linear system ofnequations

in nunknowns and ifA1exists, then

X=A1B

is the unique solution of the system.

Example

Solve the system of linear equations


141/145


Solution

Write the system of equations in the form AX=B where

2 13 2 2

2 2 1

x y zx y z

x y z

1

2

1

x

X y B

z

2 1 1

3 2 1

2 1 2

A

Example



142/145


Solution

Find the inverse matrix ofA:

2 13 2 2

2 2 1

x y zx y z

x y z

1

2

1

x

X y B

z

2 1 1

3 2 1

2 1 2

A

1

3 1 1

4 2 1

1 0 1

A

Example



143/145


Solution

Finally, we write the matrix equationX=A1

B and multiply:

2 13 2 2

2 2 1

x y zx y z

x y z

x

y

z

3 1 1 1

4 2 1 2

1 0 1 1

Example



144/145


Solution

Finally, we write the matrix equationX=A1

B and multiply:

Thus, the solution isx = 2,y =1, andz=2.

2 13 2 2

2 2 1

x y zx y z

x y z

(3)(1) ( 1)(2) ( 1)( 1) 2

( 4)(1) (2)(2) (1)( 1) 1

( 1)(1) (0)(2) (1)( 1) 2

x

y

z


145/145

End ofChapter

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LinearEqn n Matrices