OUTLINE

Matrix Operations Multiplying Matrices Determinants and Cramer’s Rule Identity and Inverse Matrices Solving systems using Inverse matrices

TYPES OF MATRICES

NAME DESCRIPTION EXAMPLE

Row matrix A matrix with only 1 row

Column matrix A matrix with only I column

Square matrix A matrix with same number of rows and columns

Zero matrix A matrix with all zero entries

3 2 1 4

23

2 4 1 7

0 00 0

MATRIX OPERATIONS

COMPARING MATRICES

For Example:

5 0

44

34

5 0 1 0.75

2 60 3

2 63 2

EQUAL MATRICES: Matrices having equal corresponding entries.

ADDING MATRICES

Matrices of same dimension can be added

For Example:

3 42

102

31 4 022

4 45

SUBTRACTING MATRICES

Matrices of same dimension can be subtracted

For example:

8 34 0

2 76 1

8 2 3 ( 7)4 6 0 ( 1)

6 10 2 1

MULTIPLYING A MATRIX BY A SCALAR

For example:

21 20 3 4 5

4 56 8 2 6

( 2)1 ( 2) 2( 2)0 ( 2)3

( 2) 4 ( 2)5

4 56 8 2 6

2 40 68 10

4 56 8 2 6

6 96 146 4

SOLVING A MATRIX EQUATION

For example:

Solve :

23x 18 5

4 1 2 y

26 012 8

23x 4 118 2 5 y

26 012 8

6x8 012 10 2y

26 012 8

Equate :6x8 26x3

10 2y8y1

MULTIPLYING MATRICES

PRODUCT OF TWO MATRICES

A3 2 1 0

B1 42 1

For example:

FIND (a.) AB and (b.) BA

SOLUTION

AB3 2 1 0

1 42 1

AB7 10 1 4

BA1 42 1

3 2 1 0

BA7 25 4

SIMPLIFY

A2 1 1 3

,B

2 04 2

,C

1 13 2

Simplify: a.) A(B+C)b.) AB+AC

SOLUTION

2 1 1 3

2 1 1 3

1 13 2

2 1 1 3

1 17 4

5 6

22 11

A(B+C):

SOLUTION

AB+AC:

2 1 1 3

2 04 2

2 1 1 3

1 13 2

0 2

14 6

5 48 5

5 6

22 11

DETERMINANTS & CRAMER”S

RULE

DETERMINANT OF 22 MATRIX

deta bc d

ad bc

The determinant of a 22 matrix is the difference of the entries on the diagonal.

EVALUATE

Find the determinant of the matrix:

1 32 5

Solution:

1 32 5

1(5) 2(3) 5 6 1

DETERMINANT OF 33 MATRIXThe determinant of a 33 matrix is the difference in the sum of the products in red from the sum of the products in black.

deta b cd e fg h i

a b cd e fg h i

a bd eg h

Determinant = [a(ei)+b(fg)+c(dh)]-[g(ec)+h(fa)+i(db)]

EVALUATE

2 1 3 2 0 11 2 4

2 1 3 2 0 11 2 4

2 1 2 01 2

[0 ( 1) ( 12)] (04 8) 13 12 25

Solution:

USING MATRICES IN REAL LIFE

The Bermuda Triangle is a large trianglular region in the Atlantic ocean. Many ships and airplanes have been lost in this region. The triangle is formed by imaginary lines connecting Bermuda, Puerto Rico, and Miami, Florida. Use a determinant to estimate the area of the Bermuda Triangle.

EW

N

S

Miami (0,0)

Bermuda (938,454)

Puerto Rico (900,-518)

. ..

SOLUTION

The approximate coordinates of the Bermuda Triangle’s three vertices are: (938,454), (900,-518), and (0,0). So the area of the region is as follows:

Area 12

938 454 1900 518 10 0 1

Area 12

[( 458,884 00) (00408,600)]

Area 447,242

Hence, area of the Bermuda Triangle is about 447,000 square miles.

USING MATRICES IN REAL LIFE

The Golden Triangle is a large triangular region in the India.The Taj Mahal is one of the many wonders that lie within the boundaries of this triangle. The triangle is formed by the imaginary lines that connect the cities of New Delhi, Jaipur, and Agra. Use a determinant to estimate the area of the Golden Triangle. The coordinates given are measured in miles.

EW

N

S

Jaipur (0,0)

New Delhi (100,120)

Agra (140,20)

. ..

SOLUTION

The approximate coordinates of the Golden Triangle’s three vertices are: (100,120), (140,20), and (0,0). So the area of the region is as follows:

Area12

100 120 1140 20 1

0 0 1

Area12

[(200000) (0016800)]

Area7400

Hence, area of the Golden Triangle is about 7400 square miles.

USING MATRICES IN REAL LIFE

Black neck stilts are birds that live throughout Florida and surrounding areas but breed mostly in the triangular region shown on the map. Use a determinant to estimate the area of this region. The coordinates given are measured in miles.

EW

N

S

(0,0)

(35,220)

(112,56). ..

SOLUTION

The approximate coordinates of the Golden Triangle’s three vertices are: (35,220), (112,56), and (0,0). So the area of the region is as follows:

Area12

35 220 1112 56 1

0 0 1

Area12

[(196000) (0024640)]

Area11340

Hence, area of the region is about 11340 square miles.

GRAPHIC USES OF MATRIX MATHEMATICSGraphic software uses matrix mathematics to process

linear transformations to render images. A square matrix, one with exactly as many rows as columns, can represent a linear transformation of a geometric object. For example, in the Cartesian X-Y plane, the matrix       reflects an object in the vertical Y axis. In a video game, this would render the upside-down mirror image of a castle reflected in a lake.If the video game has curved reflecting surfaces, such as a shiny silver goblet, the linear transformation matrix would be more complicated, to stretch or shrink the reflection.

MATRICES APPLIED TO ELECTRIC CIRCUITS

There are two closed loops in the above circuit. loop 1: e1, R1 and R3 and loop 2: e2, R2 and R3. e1 and e2 are sources of voltages. R1, R2 and R3 are resistors. i1 is the current flowing across R1 and i2 is the current flowing across R2. We now apply Kirchhoff's law to each loop. 

loop 1: e1 = R1 i1 + R3 (i1 - i2) 

loop 2: e2 = R2 i2 + R3 (i2 - i1)

Question: If e1, e2, R1, R2 and R3 are known, how do you calculate i1 and i2? This circuit is simple and involves only two equations. However electric circuits can be much more complicated that the one above and matrices are suitable to answer the above question. Let us group like terms in the above system of equations 

e1 = i1 (R1 + R3) - i2 R3 

e2 = - i1 R3 + i2(R2 + R3) 

and then write it in matrix form as follows 

The above is a matrix equation that may be solved using any known method to solve systems of equations. Let e, R and i be matrices given by 

The solution to the above matrix equation is given by

where R -1 is the inverse matrix of R and is given by. 

USING MATRICES

Example Let the message be 

PREPARE TO NEGOTIATE  and  the encoding matrix be

We assign a number for each letter of the alphabet. For simplicity, let us associate each letter with its position in the alphabet: A is 1, B is 2, and so on. Also, we assign the number 27 (remember we have only 26 letters in the alphabet) to a space between two words. Thus the message becomes:

 Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1 vectors:

 Note that it was necessary to add a space at the end of the message to complete the last vector. We now encode the message by multiplying each of the above vectors by the encoding matrix. This can be done by writing the above vectors as columns of a matrix and perform the matrix multiplication of that matrix with the encoding matrix as follows:  

which gives the matrix

The columns of this matrix give the encoded message. The message is transmitted in the following linear form

 To decode the message, the receiver writes this string as a sequence of 3 by 1 column matrices and repeats the technique using the inverse of the encoding matrix. The inverse of this encoding matrix, the decoding matrix, is: 

(make sure that you compute it yourself). Thus, to decode the message, perform the matrix multiplication

and get the matrix

 The columns of this matrix, written in linear form, give the original message:

CRAMER”S RULE FOR A 22 SYSTEM

Let A be the co-efficient matrix of the linear system: ax+by= e & cx+dy= f.

IF det A ≠0, then the system has exactly one solution. The solution is:

x

e bf ddetA

y

a ec fdetA

The numerators for x and y are the determinant of the matrices formed by using the column of constants as replacements for the coefficients of x and y, respectively.

EXAMPLE

Use cramer’s rule to solve this system:

8x+5y = 2 2x-4y = -10

SOLUTION

Solution: Evaluate the determinant of the coefficient matrix

8 52 4

32 10 42

Apply cramer’s rule since the determinant is not zero.

x

2 5 10 4

42

8 ( 50) 42

42 42

1

y

8 22 10 42

80 4 42

84 42

2 The solution is (-1,2)

CRAMER”S RULE FOR A 33 SYSTEM

Let A be the co-efficient matrix of the linear system: ax+by+cz= j, dx+ey+fz= k, and gx+hy+iz=l.

IF det A ≠0, then the system has exactly one solution. The solution is:

x

j b ck e fl h i

detA, y

a j cd k fg l i

detA, z

a b jd e kg h l

detA

EXAMPLE

The atomic weights of three compounds are shown. Use a linear system and Cramer’s rule to find the atomic weights of carbon(C ), hydrogen(H), and oxygen(O).

Compound Formula Atomic weight

Methane CH4 16

Glycerol C3H8O3 92

Water H2O 18

SOLUTION

1 4 03 8 30 2 1

(800) (0612) 10

Write a linear system using the formula for each compound

C + 4H = 163C+ 8H + 3O = 92 2H + O =18

Evaluate the determinant of the coefficient matrix.

SOLUTION

Apply cramer’s rule since determinant is not zero.

C

16 4 092 8 318 2 1

10

120 10

12

H

1 16 03 92 30 18 1

10

10 10

1

O

1 4 163 8 920 2 18

10

160 10

16

Atomic weight of carbon = 12

Atomic weight of hydrogen =1

Atomic weight of oxygen =16

IDENTITY AND INVERSE

MATRICES

IDENTITY MATIX

22 IDENTITY MATRIX 33 IDENTITY MATRIX

I 1 00 1

I 1 0 00 1 00 0 1

INVERSE MATRIX

The inverse of the matrix

Aa bc d

is

A 1 1A

d b c a

A 1 1

ad cbd b c a

providedad cb0

EXAMPLE

Find the inverse of

A3 14 2

Solution:

A 1 1

6 42 1 4 3

12

2 1 4 3

1 12

232

CHECK THE SOLUTION

Show

AA 1 I A 1A

3 14 2

1 12

232

1 00 1

,

and

1 12

232

3 14 2

1 00 1

SOLVING SYSTEMS USING INVERSE

MATRICES

SOLVING A LINEAR SYSTEM

-3x + 4y = 5 2x - y = -10

Writing the original matrix equation.

3 42 1

xy

5 10

A X B AX = BA-1AX = A-1B IX = A-1B X = A-1B

USING INVERSE MATRIX TO SOLVE THE LINEAR SYSTEM-3x + 4y = 5

2x - y = -10

A 1 1

3 8 1 4 2 3

15

45

25

35

X A 1B

15

45

25

35

5 10

7 4

xy

Hence the solution of the system is (-7,-4)