MATH 1431-Precalculus Imathlady.org/mathlady/MATH1431 Ch 5 Systems and Matrices COD Spring... ·...

MATH 1431-Precalculus I

Chapter 5-Sytems and Matrices

Solving Systems of Two Linear Equations

Two-Variable Linear Equations

A system of equations is composed of two or more equations considered

simultaneously. The solution set of the system consists of all ordered

pairs that make both equations true.

I. Solving Systems of Equations Graphically

A. The system has one solution if exactly one point of intersection exists in

a system of two linear equations.

B. There is no solution between two parallel lines (no common points).

C. There are infinitely many solutions if two lines described are identical...

same line (infinitely many common points).

D. If a system of two linear equations in two variables has one solution, it is

consistent.

E. If a system has no solution, it is inconsistent.

F. If a system of two linear equations in two variables has infinite number of solutions, it is dependent. Otherwise, it is independent.

Example:

Consider the system: 2 2x y+ = , x y- = 7

1

Math1431 S. Nunamaker

Consider the system: x y+ = -2, y x= - 8

II. Solving Systems of Equations Using the Substitution

Example: Solve the system x y+ = 11 , 3 5x y- =

First, consider the x y+ = 11 equation and solve for y.

y x= -11

Then, consider the 3 5x y- = equation and replace

y by the 11 - x.

So, 3 5x y- = becomes 3 11 5x x- - =( ) .

So, 3 11 5x x- - =( ) becomes 3 11 5x x- + = .

So, 3 11 5x x- + = becomes 4 11 5x - = or 4 16x = ..

So, x = 4.

Now, substitute 4 for x in either x y+ = 11 or

3 5x y- = , the result will be y = 7.

2

MATH1431 S. Nunamaker

Example: Solve the system x y- =5 4, y x= -7 2 .

III. Solving Systems of Equations Using Elimination

Example: Solve the System 2 2x y+ = , x y- = 7

Try to eliminate one of the two variables by :

1. adding two equations if one of the two

variables have same coefficients and

opposite signs in two equations.

OR 2. subtracting two equations if one of the two

variables have same coefficients and same

signs in two equations.

So, 2 2x y+ =

+ ( )x y- = 7

3 0 9x + = , so x = 3

Then, use back-substitution to solve for y, into

x y- = 7, so 3 7- =y , so y = -4.

3


Example: Solve the system x y- =3 1, - + =2 6 5x y

Solve the system: 5 7 23

3 2 11

x y

x y

- =

+ = -

Solve the system:

5

4

2

51

1

4

2

311

x y

x y

- = -

+ =

4


Determinants

The Determinant of a Square Matrix

I. Determinant of a 2 2X matrix:

if A A=-é

ëê

ù

ûú =

-= - - = + =

2 3

1 2

2 3

1 22 2 1 3 4 3 7,det( ) ( ) ( )

if B B=é

ëê

ù

ûú = = - = - =

2 1

4 2

2 1

4 22 2 4 1 4 4 0,det( ) ( ) ( )

II. Determinant of a 3 3X or higher order matrix:

It is convenient to introduce the concepts of minors and cofactors.

A. If A is a square matric, the minor M ij of the entry aij is the

determinant of the matrix obtained by deleting the ith row and jth column of A.

The cofactor C ij of the entry aij is given by C Miji j

ij= - +( )1

Example: Find all the minors and cofactors of A = -

é

ë

êêê

ù

û

úúú

0 2 1

3 1 2

4 0 1

To find the minor M11

1 2

0 11 1 0 2 1=

-= - - = -( ) ( ) (by deleting the first row

and first column of A. and evaluate the determinant of the resulting matrix.)

M12

3 2

4 13 1 4 2 3 8 5= = - = - = -( ) ( ) , continuing this pattern,

M M M M M M11 12 13 21 22 231 5 4 2 4 8= - = - = = = - = -, , , , ,

M M M31 32 335 3 6= = - = -, ,

To find cofactors: C M111 1

1121 1 1 1= - = - - = -+( ) ( ) ( )

5


C C C C C C11 12 13 21 22 231 5 4 2 4 8= - = = = - = - =, , , , , , , ,C C C31 32 335 3 6= = = -

III. Determinant of a Square Matrix

If A is a square matrix ( of order 2x2 or greater ), the determinant of A isthe sum of the entries in any row (or column ) of A multiplied by their respectivecofactors. For instance, expanding along the first row yields

A a C a C a Cn n= + + +11 11 12 12 1 1...

applying this definition to find a determinant is called expanding by cofactors.

For a 2x2 matrix, this definition yields A a a a a= -11 22 12 21

For a 3x3 matrix, A = -

é

ë

êêê

ù

û

úúú

0 2 1

3 1 2

4 0 1

note that C C C11 12 131 5 4= - = =, ,

\determinant of A, A= + +a C a C a C11 11 12 12 13 13= - + + =0 1 2 5 1 4 14( ) ( ) ( )

or, A a C a C a C= + + = - + - - + =21 21 22 22 23 23 3 2 1 4 2 8 14( ) ( )( ) ( )

Example: Find the determinant of -

-

é

ëê

ù

ûú

3 4

2 1

Find the determinant of

6 3 7

0 0 0

4 6 3

-

-

é

ë

êêê

ù

û

úúú

Find the determinant of

-

-

é

ë

êêê

ù

û

úúú

1 3 1

4 2 5

2 1 6

6


Properties of Determinants

I. If every element in a row (or a column) of a determinant is zero, the value

of the determinant is zero. ex.:

3 2 6

0 0 0

1 5 8

0

5 3 0

7 2 0

8 6 0

0

-

-

= -

-

=,

II. If two rows (or two columns) of a determinant are identical, the value of

the determinant is zero. ex.:

1 3 5

2 6 3

1 3 5

0

6 2 2

0 5 5

2 3 3

0

-

- -

-

=

- -

=,

III. If any two rows (or two columns) of a determinant are interchanged, the

sign of the value of the determinant is changed. ex. :

3 2 7

5 8 4

0 6 9

5 8 4

3 2 7

0 6 9

-

-

= - -

-

ex. :

0 3 5

6 2 3

9 7 1

5 3 0

3 2 6

1 7 9

-

- = -

-

-

IV. If every element of a row (or a column) is multiplied by the same realnumber k, the value of the determinant is multiplied by k. ex.: 3 5 6

4 8 12

3 0 7

4

3 5 6

1 2 3

3 0 7

-

- =

-

-

ex.: 3

1 4 6

3 0 8

2 4 3

1 4 18

3 0 24

2 4 9

-

-

=

-

-

7


V. If every element of a row (or a column) is multiplied by the same realnumber k, and if the resulting products are added to another row (or anothercolumn), the value of the determinant remains the same.

ex.: evaluate

3 2 6

6 0 4

5 1 1

2

13 0 4

6 0 4

5 1 13 1 1

-

-

-

+ ® ® -

-

( )R R R

expanding down the second column, we find that the first two minorsresult in zero.

Therefore,

13 0 4

6 0 4

5 1 1

113 4

6 41 52 24 76-

-

= --

= - - - = -[ ( )]

if we evaluate this determinant by expansion by minors, we obtain thesame result:

3 2 6

6 0 4

5 1 1

-

-

-

let's expand by minors down the second column as follows:

- --

-+

--

-( )2

6 4

5 10

3 6

5 11

3 6

6 4

= -2 6 20( ) + - -0 3 30( )- +1 12 36( )

= -28 +0 -48 = -76

Examples: 1. Evaluate each determinant:

a.

4 3 8

2 7 9

0 0 0

-

- b.

- -3 6 3

1 5 1

7 0 7

c.

3 1 6

4 0 7

5 6 2

-

-

8


Solving A System of Linear Equations Using Determinants

I. The solution of a x b y c

a x b y c1 1 1

2 2 2

+ =

+ = may be written in determinant form as

follows:

x

c b

c b

a b

a b

=

1 1

2 2

1 1

2 2

y

a c

a c

a b

a b

=

1 1

2 2

1 1

2 2

*If the determinant of the numerator is not zero and determinant of thedenominator is zero, the system is inconsistent. If the determinant of numeratorand denominator are both zero, the system is dependent. If the determinant of thedenominator is not zero, there is a unique solution and the system is independentand consistent.

II. Cramer's Rule: To solve a system of two linear equations in twovariables.

* To find the determinant of the numerator:

a. For x, take the determinant of the denominator and replace the

coefficients of x, a s' , by the corresponding constants, c s' .

b. For y, take the determinant of the denominator and replace the

coefficients of y b s, ' , by the corresponding constants, c s' .

III. Examples:

Try solving this system of linear equations, using determinants:

A. 2 3 22

5 4 14

x y

x y

- =

+ = -

B. 3 4 10

6 8 5

x y

x y

+ =

+ = -

9


IV. To solve a system of three linear equations with three variables usingdeterminants:

A. The general system of three equations in three variables is given as

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1 1

2 2 2 2

3 3 3 3

+ + =

+ + =

+ + =

The general solution may be written in terms of determinants as follows:

x

d b c

d b c

d b c

a b c

a b c

a b c

=

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

y

a d c

a d c

a d c

a b c

a b c

a b c

=

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

z

a b d

a b d

a b d

a b c

a b c

a b c

=

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

B. Examples:

Try to solve:

.

3 4 15

2 5 29

6 24

x y z

x y

x y z

- + = -

+ =

- - = -

10


V. Solve a system of equations with a unique solution, using inverse of matrix:

If A is an invertible matrix, the system of linear equations represented by AX B= has a unique solution given by X A B= -1

Example: Use an inverse matrix to solve this system of equations:

2 3 1

3 3 1

2 4 2

x y z

x y z

x y z

+ + = -

+ + =

+ + = -

So, A B=

é

ë

êêê

ù

û

úúú

=

-é

ë

êêê

ù

û

úúú

2 3 1

3 3 1

2 4 1

1

1

2

,

A- =

-

-

- -

é

ë

êêê

ù

û

úúú

1

1 1 0

1 0 1

6 2 3

\ solution X =

-

-

- -

é

ë

êêê

ù

û

úúú

-é

ë

êêê

ù

û

úúú

= -

-

é

ë

êêê

1 1 0

1 0 1

6 2 3

1

1

2

2

1

2

ù

û

úúú

\ the solution is: x y z= = - = -2 1 2, ,

Example: Find the solution of:

3 2 2 0

2 2 2 5

4 4 3 2

x y z

x y z

x y z

+ + =

+ + =

- + + =

11


Matrices

For linear systems involving three or more variables, the technique of usinga rectangular array of real numbers is called a matrix (the plural form of matrix ismatrices).

I. Definitions

The system 2 3 7x y- = , x y+ = -4 2 can be written as

2 3 7

1 4 2

-

- This rectangular array of numbers is called

a matrix, more specifically an augmented matrix.

2 3

1 4

- is the coefficient matrix of the system.

rows: (horizontal)

column: (vertical)

order: size; m x n

square matrix: m = n;number of rows = number of columns

main diagonal elements/entries: elements/entries at the

location where position of row = position of column.

Augmented Matrix:: For a system of equations, it is a table of numbers en-

closed in brackets, where the rows represent the equations, all columns, but

the last column hold the coefficients of the variables in the equations, and

the last column is the right side of the equations.

12


Ex. x y z+ - = -2 3

- - + =x y z 0

1 1 2 3

1 1 1 0

2 3 2 2

- -

- -

-

2 3 2 2x y z+ - =

*The main diagonal of an augmented matrix is the set of all augmented matrix

entries that lie to the left of the vertical line and have identical row and

column positions.

*An augmented matrix is upper triangular if all entries below the main dia-

gonal are zero. A system of equations is ready for solution by back-substi-

tution when its augmented matrix is upper triangular and when all nonzero

elements on the main diagonal are ones.

Ex.

1 2 3 11

0 4 5 12

0 0 6 13

Main diagonal elements are: 1, 4, 6

Ex. 1 0 2 0 8

0 0 1 3 0 Main diagonal elements are: 1, 0

Ex.

1 2 6 24 4

0 1 4 9 6

0 0 1 2 857

.

.

.

This augmented matrix is Upper Triangular

13


II. Elementary Row Operations: Operations on equations that change the look

of the equations without changing the solution of the system.

A. Interchange the positions of any two equations.

B. Multiply an equation by a nonzero number.

C. Add to one equation a nonzero constant times another equation.

III. Gaussian Elimination

The process of using elementary row operations to place zeros below the

main diagonal of an augmented matrix.

A. Pivot is a nonzero element in an augmented matrix that is used to

transform elements below it to zero. Pivot must be nonzero. If an

element on the main diagonal is zero when we need it as a pivot, we use

the first elementary row operation to interchange the row containing the

zero diagonal entry with any

row below it that has a nonzero entry in the same column as the zero diago-

nal entry.

Ex. 0 2 4

1 3 1- ®

1 3 1

0 2 4

- by interchanging the first and second row

1 3 1

0 2 4

- ®

1 3 1

0 1 2

- by multiplying the second row by 1/2

1/2 R R2 2®

14


IV. Gauss-Jordan Reduction:

Step 1: Use Gaussian elimination to transform an augmented matrix into

the augmented matrix that is upper triangular matrix with all

nonzero elements on the main diagonal set to one.

Step 2: Beginning with the last column of the upper triangular matrix

having a one on its main diagonal and progressing backward se-

quentially to the second column, use only the third elementary

row operation to transform all elements in the upper triangular

matrix above the main diagonal to zero. Complete all work on

column before moving to another column, and apply all operations

to the entire augmented matrix.

Ex. x + y - 2z = -3 x y z

-x - y + z = 0

1 1 2 3

1 1 1 0

2 3 2 2

- -

- -

-

2x + 3y - 2z = 2 R R R2 1 2+ ®

1 1 2 3

0 0 1 3

2 3 2 2

- -

- -

-

R R R3 1 32+ - ®( )

1 1 2 3

0 0 1 3

0 1 2 8

- -

- -

15


R R2 3«

1 1 2 3

0 1 2 8

0 0 1 3

- -

- -

( )- ®1 3 3R R

This is in the upper triangular matrix form:

1 1 2 3

0 1 2 8

0 0 1 3

- -

This completes Step 1 of Gauss-Jordan reduction. We now begin with

column 3 and use elementary row operations to place zeros above the one on

main diagonal. Once this is done, we move to column 2 and place a zero

above the one on the main diagonal in that column. We have:

1 1 2 3

0 1 2 8

0 0 1 3

- -

R R R2 3 22+ - ®( ) ®

1 1 2 3

0 1 0 2

0 0 1 3

- -

Which eliminates z from the second equation, and

R R R1 3 12+ ®

1 1 0 3

0 1 0 2

0 0 1 3

Which eliminates z from the first equation.

R R R1 2 11+ - ®( )

16


1 0 0 1

0 1 0 2

0 0 1 3

The last step eliminates y from the first equation. The system of equations

corresponding to this augmented matrix is x = 1, y = 2, z = 3, which also

gives the solution to the original system without requiring any back-substi-

tution.

*If one of the equations associated with the augmented matrix in upper

triangular form is false, then a system of linear equations has no solution.

Example: Solve the system:

2 4 3x y z- + = -

x y z- - = -2 10 6

3 4 7x z+ =

The augmented matrix for this system:

2 1 4 3

1 2 10 6

3 0 4 7

- -

- - -

The goal is to find a row-equivalent matrix of the form:

1

0 1

0 0 1

a b c

d e

f

17


2 1 4 3

1 2 10 6

3 0 4 7

- -

- - - -----R R1 2« ---->

1 2 10 6

2 1 4 3

3 0 4 7

- - -

- -

1 2 10 6

2 1 4 3

3 0 4 7

- - -

- - -- - + ®2 1 2 2R R R -->

1 2 10 6

0 3 24 9

3 0 4 7

- - -

1 2 10 6

0 3 24 9

3 0 4 7

- - -

--- + ®3 1 3 3R R R -->

1 2 10 6

0 3 24 9

0 6 34 25

- - -

1 2 10 6

0 3 24 9

0 6 34 25

- - -

--- 1

32 2R R® --->

1 2 10 6

0 1 8 3

0 6 34 25

- - -

*Work on elements in each column at a time, starting

from the element/entry in the main diagonal, go .

1 2 10 6

0 1 8 3

0 6 34 25

- - -

- - + ®6 2 3 3R R R --->

1 2 10 6

0 1 8 3

0 0 14 7

- - -

-

18


1 2 10 6

0 1 8 3

0 0 14 7

- - -

-

-- -

®1

143 3R R --->

1 2 10 6

0 1 8 3

0 0 11

2

- - -

-

At this point, the system of equations that corresponds

to the last matrix above is: x y z- - = -2 10 6 (1)

y z+ =8 3 (2)

z = -1

2 (3)

Now, we may back-substitute -1

2 for z in equation (2)

and solve for y: y + - =81

23( )

y - =4 3

y = 7

Now, we back-substitute 7 for y and -1

2 for z in

equation (1 ) and solve for x: x - × - - = -2 7 101

26( )

x - + = -14 5 6

x - = -9 6

x = 3

*The last matrix is in row-echelon form.

19


V. Row-Echelon Form

To be in this form, a matrix must have the following properties:

1. If a row does not consist entirely of 0's, then the first

nonzero element in the row is a 1 (called a leading 1).

2. For any two successive nonzero rows, the leading 1 in

the lower row is farther to the right than the leading 1

in the higher row.

3. All the rows consisting entirely of 0's are at the bottom

of the matrix.

If a fourth property is also satisfied, a matrix is said to be

in reduced row-echelon form:

4. Each column that contains a leading 1 has 0's every-

where else (if every column that has a leading 1 has zeros in

every position above and below its leading 1.

VI. Gauss-Jordan Elimination

This method is named for Karl Friedrich Gauss and Wilhelm

Jordan (1842-1899). This is the continuation steps following

the Gaussian Elimination (as a an alternative to back-

substitution ) to solve the system of equations.

Using previous example, at the end of Gaussian Elimination,

20


we have:

1 2 10 6

0 1 8 3

0 0 11

2

- - -

-

We need to continue performing row-equivalent

operations until we have a matrix in reduced row-echelon

form. We need to work from the third column first, and

then the second column.

1 2 10 6

0 1 8 3

0 0 11

2

- - -

-

-- - + ®8 3 2 2R R R --->

1 2 10 6

0 1 0 7

0 0 11

2

- - -

-

1 2 10 6

0 1 0 7

0 0 11

2

- - -

--- 10 3 1 1R R R+ ® -->

1 2 0 11

0 1 0 7

0 0 11

2

- -

-

1 2 0 11

0 1 0 7

0 0 11

2

- -

-

-- 2 2 1 1R R R+ ® --->

1 0 0 3

0 1 0 7

0 0 11

2-

At this point, we have x = 3, y = 7, z = -1

2 read directly

from the reduced row-echelon matrix.

21


Operations with Matrices

A. A matrix is a rectangular array of elements arranged in horizontal rowsand vertical columns.

Ex. A =

3 4 5

2 7 6

3 0 3-

B = 5 9 6 0

23 50 7 200

B. Column vectors are matrices having a single column, such as:

Ex. a = 6

4 b =

100

2

4

c =

2

4

5

9

d =

0

9

6

2

10

C. Row vectors are matrices having a single row, such as:

Ex. g = 3 7 h = 9 11 7 p = 6 10 7 0

*Uppercase boldface type for most matrices and lowercase boldface

type for row vectors and column vectors.

D. Number of rows is specified before the number of columns when

giving the size or order of a matrix. For example, as listed above

the order or size of these matrices are:

A is a 3 x 3 (three by three), B is a 2 x 4 (two by four), a is a 2 x 1

(two by one), b is a 3 x 1 (three by one), c is a 4 x 1 (four by

one), d is a 5 x 1 (five by one), g is a 1 x 2 (one by two), h is a 1 x 3

(one by three), p is a 1 x 4 (one by four)

22


*Since row vectors have only one row and column vectors have only

one column, it is common to specify their sizes just by listing the

number of columns for a row vector or the number of rows for a

column vector. This number is the dimension of a vector.

For example, b is a 3-dimensional column vector while p is a4-dimensional

row vector.

E. Elements in a matrix are identified by a lowercase letter and two

subscripts; the first subscript denotes the row position of the element

and the second subscript denotes the column position of the element.

Ex. a23reads "a sub two three", represents the element of A located

in second row and third column.

F. Two matrices are equal if they have the same order/size and if their

corresponding elements are equal.

G. The sum of two matrices of the same order is the matrix obtained

by adding together the corresponding elements of the original

matrices..

Ex. if A =

2 3 4

1 2 0

4 7 1

B =

0 1 2

2 3 1

4 3 2

, then A + B =

2 4 6

3 5 1

8 10 3

23


H. A zero matrix, 0, is a matrix whose elements are all equal to zero.

Ex.

0 0 0

0 0 0

0 0 0

I. The difference A - B of two matrices of the same order is the

matrix obtained by subtracting from the elements of A

corresponding elements of B.

if A =

2 3 4

1 2 0

4 7 1

B =

0 1 2

2 3 1

4 3 2

,. then A - B =

2 2 2

1 1 1

0 4 1

- - -

-

J. The product of a number c and a matrix A is the matrix obtained

by every element of A by c.

Ex. if B =- -1 0 1

0 2 1 2/ then -3B=

3 0 3

0 6 3 2- - /

K. The product of two matrices AB is defined if and only if the

number of columns in A equals the number of rows in B.

L. To calculate the i-j elements of AB, multiply the element in the

ith row of A by the corresponding elements in the jth column of

B and sum the results.

For example, if A = 0 1 2

3 4 5 and B =

6 7

8 9

1 2- -

= 11 12

21 22

24


the 1-1 element (i=1, j=1) is obtained by multiplying the elements in the

first row of A by the corresponding elements in the first column of B

and summing the result. 0 1 2

6

8

1-

= 0 (6) + 1 (8) +2 (-1)= 6

the 1-2 element (i-1, j=2) is obtained by multiplying the elements in the

first row of A by the corresponding elements in the second column of B

and summing the result.0 1 2

7

9

2-

= 0 (7) + 1 (9) + 2 (-2)= 5

the 2-1 element (i=2, j=1) is obtained by multiplying the elements in the

second row of A by the corresponding elements in the first column of B

and summing the result. 3 4 5

6

8

1-

= 3 (6) + 4 (8) + 5 (-1)= 45

the 2-2 element (i=2, j=2) is obtained by multiplying the elements in

the second row of A by the corresponding elements in the second

second column of B and summing the results.

3 4 5

7

9

2-

= 3 (7) + 4 (9) + 5 (-2) = 47

25


So, we have AB = 0 1 2

3 4 5

6 7

8 9

1 2- -

= 6 5

45 47

M. Properties of Matrix Multiplication

Let A, B, and C be matrices and let c be a scalar.

1. A BC AB C( ) ( )= 2. A B C AB AC( )+ = +

3. ( )A B C AC BC+ = + 4. c AB cA B A cB( ) ( ) ( )= =

Examples: Given [ ]A B C D E=é

ëê

ù

ûú =

é

ëê

ù

ûú = - =

é

ëê

ù

ûú =

-

2 3

6 1

4

52 1

1 0

0 1

3 6

2 1, , , ,

é

ëê

ù

ûú

Determine: A B A D A E AB AC AD AE+ + -, , , , , , ,

3( ),( ) , ( ),( ) ,AB AB C A D E A E D+ +

N.. Identity Matrix, I

This is a square matrix in which all main diagonal elements are

equal to one and all other elements are equal to zero.

Ex. 1 0

0 1 ,

1 0 0

0 1 0

0 0 1

,

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

,

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

26


If A and I have compatible orders so that their product is defined

(same number of column from A as that of the number of row from I),

then AI = A and IA = A

Matrix Inversion/Inverse of a Square Matrix

I. Every system of linear equations can be written in the form of the

matrix equation Ax = B, such that A is a coefficient matrix, x is

a vector of variables, and b is a constant vector.

Ex. To write the system of equations: x + 2y + 3z = 10

4x - 5y = 20

in the matrix form Ax = b

A = 1 2 3

4 5 0- , x =

x

y

z

, b = 10

20

So, the original system of equations can be written as the matrix equation:

1 2 3

4 5 0-

x

y

z

= 10

20

II. A matrix B is the inverse of a square matrix A if and only if their

product is an identity matrix: AB = BA = I

[ or think in terms of 6 (1/6)=1, such that 1/6 is the inverse of 6 and

1 is corresponding to I]

27


Ex. To determine if B = 2 5

1 3

-

- is an inverse of A =

3 5

1 2 ,

we need to find out if the product AB is the identity matrix.

So, check to see if AB = 3 5

1 2

2 5

1 3

-

- =

1 0

0 1 = I

1-1 element is 3 5 2

1- = 3 (2) + 5 (-1)= 1

1-2 element is 3 5 -5

3 = 3 (-5) + 5 (3)= 0

2-1 element is 1 2 2

1- = 1(2) + 2 (-1) =0

2-2 element is 1 2 -5

3 = 1 (-5) + 2 (3) =1

Therefore, AB = 1 0

0 1 and is the identity matrix.

So, B is an inverse of A.

*When the inverse of a square matrix A exists, it is denoted by A-1 .

II. Inversion Algorithm

Step 1. Form the partitioned matrix A I, where I is the identity

matrix having the same order as A.

Step 2. Use elementary row operations to transform A into upper

triangular form, applying each operation to the entire parti-

tioned matrix. Denote the result as C D , where C is in upper

28


triangular form with one as the first nonzero element in each

nonzero row.

Step 3. Check whether C has any zeros on its main diagonal. If it

does,stop; A does not have an inverse. Otherwise, continue.

Step 4. Beginning with the last column of C and progressing

backward sequentially to the second column, use only the

third elementary row operation to transform all elements

above the main diagonal of C to zero. Complete all work on

one column before moving on to the next column, and apply

all operations to the entire partitioned matrix.

Step 5. At the conclusion of Step 4, the partitioned matrix will have

the form I B , with B = A-1 .

III. If A has an inverse, then the unique solution to the matrix equation

Ax = b is x = A b-1

IV. The Inverse of a 2 2X Matrix (Quick Method )

This only works for a 2 2X matrix: if Aa b

c d=

é

ëê

ù

ûú

then A is invertible if and only if ad bc- ¹ 0, then

Aad bc

d b

c a- =

-

-

-

é

ëê

ù

ûú

1 1

*ad bc- is called the determinant of the 2 2X matrix A

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Example: if

A A=-

-

é

ëê

ù

ûú \ =

× - - -

é

ëê

ù

ûú =-

3 1

2 2

1

3 2 1 2

2 1

2 3

1

4

2 1

2 31,

( )( )

é

ëê

ù

ûú =

é

ë

êêê

ù

û

úúú

1

2

1

41

2

3

4

V. A System of Equations with a Unique Solution

If A is an invertible matrix, the system of linear equations represented by AX B= has a unique solution given by X A B= -1

Example: Use an inverse matrix to solve this system of equations:

2 3 1

3 3 1

2 4 2

x y z

x y z

x y z

+ + = -

+ + =

+ + = -

So, A B=

é

ë

êêê

ù

û

úúú

=

-é

ë

êêê

ù

û

úúú

2 3 1

3 3 1

2 4 1

1

1

2

,

A- =

-

-

- -

é

ë

êêê

ù

û

úúú

1

1 1 0

1 0 1

6 2 3

\ solution X =

-

-

- -

é

ë

êêê

ù

û

úúú

-é

ë

êêê

ù

û

úúú

= -

-

é

ë

êêê

1 1 0

1 0 1

6 2 3

1

1

2

2

1

2

ù

û

úúú

\ the solution is: x y z= = - = -2 1 2, ,

Example: Find the solution of:

3 2 2 0

2 2 2 5

4 4 3 2

x y z

x y z

x y z

+ + =

+ + =

- + + =

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MATH 1431-Precalculus Imathlady.org/mathlady/MATH1431 Ch 5 Systems and Matrices COD Spring... ·...

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