MATRICES & DETERMINANTS

Monika V Sikand

Light and Life Laboratory

Department of Physics and Engineering physics

Stevens Institute of Technology

Hoboken, New Jersey, 07030.

OUTLINE

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

MATRIX

A rectangular arrangement of numbers in rows and columns

For example:

€

6 2 −1

−2 0 5

⎡

⎣ ⎢

⎤

⎦ ⎥2 rows

3 columns

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[ ]

€

2

3

⎡

⎣ ⎢

⎤

⎦ ⎥

€

2 4

−1 7

⎡

⎣ ⎢

⎤

⎦ ⎥

€

0 0

0 0

⎡

⎣ ⎢

⎤

⎦ ⎥

MATRIX OPERATIONS

COMPARING MATRICES

For Example:

€

5 0

−4

4

3

4

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥=

5 0

−1 0.75

⎡

⎣ ⎢

⎤

⎦ ⎥

€

−2 6

0 −3

⎡

⎣ ⎢

⎤

⎦ ⎥≠

−2 6

3 −2

⎡

⎣ ⎢

⎤

⎦ ⎥

EQUAL MATRICES: Matrices having equal corresponding entries.

ADDING MATRICES

Matrices of same dimension can be added

For Example:

€

3

−4

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

1

0

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

3+1

−4 + 0

2 + 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

4

−4

5

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

SUBTRACTING MATRICES

Matrices of same dimension can be subtracted

For example:

€

8 3

4 0

⎡

⎣ ⎢

⎤

⎦ ⎥−

2 −7

6 −1

⎡

⎣ ⎢

⎤

⎦ ⎥=

8 − 2 3−(−7)

4 − 6 0 −(−1)

⎡

⎣ ⎢

⎤

⎦ ⎥=

6 10

−2 1

⎡

⎣ ⎢

⎤

⎦ ⎥

MULTIPLYING A MATRIX BY A SCALAR

For example:

€

−2

1 −2

0 3

−4 5

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

−4 5

6 −8

−2 6

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

(−2)1 (−2) − 2

(−2)0 (−2)3

(−2) − 4 (−2)5

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

−4 5

6 −8

−2 6

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=

−2 4

0 −6

8 −10

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥+

−4 5

6 −8

−2 6

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=

−6 9

6 −14

6 −4

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

SOLVING A MATRIX EQUATION

For example:

€

Solve :

23x −1

8 5

⎡

⎣ ⎢

⎤

⎦ ⎥+

4 1

−2 −y

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟=

26 0

12 8

⎡

⎣ ⎢

⎤

⎦ ⎥

23x+ 4 −1+1

8 − 2 5 − y

⎡

⎣ ⎢

⎤

⎦ ⎥=

26 0

12 8

⎡

⎣ ⎢

⎤

⎦ ⎥

6x+ 8 0

12 10 − 2y

⎡

⎣ ⎢

⎤

⎦ ⎥=

26 0

12 8

⎡

⎣ ⎢

⎤

⎦ ⎥

Equate :

6x+ 8 = 26

x = 3

10 − 2y = 8

y =1

MULTIPLYING MATRICES

PRODUCT OF TWO MATRICES

€

A =3 2

−1 0

⎡

⎣ ⎢

⎤

⎦ ⎥

B =1 −4

2 1

⎡

⎣ ⎢

⎤

⎦ ⎥

For example:

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

SOLUTION

€

AB =3 2

−1 0

⎡

⎣ ⎢

⎤

⎦ ⎥1 4

2 1

⎡

⎣ ⎢

⎤

⎦ ⎥

AB =7 −10

−1 4

⎡

⎣ ⎢

⎤

⎦ ⎥

BA =1 −4

2 1

⎡

⎣ ⎢

⎤

⎦ ⎥3 2

−1 0

⎡

⎣ ⎢

⎤

⎦ ⎥

BA =7 2

5 4

⎡

⎣ ⎢

⎤

⎦ ⎥

SIMPLIFY

€

A =2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥,B =

−2 0

4 2

⎡

⎣ ⎢

⎤

⎦ ⎥,C =

1 1

3 2

⎡

⎣ ⎢

⎤

⎦ ⎥

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

SOLUTION

€

=2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥

2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥+

1 1

3 2

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟

=2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥−1 1

7 4

⎡

⎣ ⎢

⎤

⎦ ⎥

=5 6

22 11

⎡

⎣ ⎢

⎤

⎦ ⎥

A(B+C):

SOLUTION

AB+AC:

€

=2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥−2 0

4 2

⎡

⎣ ⎢

⎤

⎦ ⎥+

2 1

−1 3

⎡

⎣ ⎢

⎤

⎦ ⎥1 1

3 2

⎡

⎣ ⎢

⎤

⎦ ⎥

=0 2

14 6

⎡

⎣ ⎢

⎤

⎦ ⎥+

5 4

8 5

⎡

⎣ ⎢

⎤

⎦ ⎥

=5 6

22 11

⎡

⎣ ⎢

⎤

⎦ ⎥

DETERMINANTS & CRAMER”S RULE

DETERMINANT OF 22 MATRIX

€

deta b

c 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 3

2 5

⎡

⎣ ⎢

⎤

⎦ ⎥

Solution:

€

1 3

2 5=1(5) − 2(3) = 5 − 6 = −1

DETERMINANT OF 33 MATRIX

The determinant of a 33 matrix is the difference in the sum of the products in red from the sum of the products in black.

€

det

a b c

d e f

g h i

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

a b c

d e f

g h i

a b

d e

g h

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

EVALUATE

€

2 −1 3

−2 0 1

1 2 4

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

2 −1 3

−2 0 1

1 2 4

2 −1

−2 0

1 2

=[0 +(−1)+ (−12)]− (0 + 4 + 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 = ±1

2

938 454 1

900 −518 1

0 0 1

Area = ±1

2[(−458,884 + 0 + 0) − (0 + 0 + 408,600)]

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:

€

Area = ±1

2

100 120 1

140 20 1

0 0 1

Area = ±1

2[(2000 + 0 + 0) − (0 + 0 +16800)]

Area = 7400

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:

€

Area = ±1

2

35 220 1

112 56 1

0 0 1

Area = ±1

2[(1960 + 0 + 0) − (0 + 0 + 24640)]

Area =11340

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

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 b

f d

detA

y =

a e

c f

detA

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 5

2 −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 2

2 −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 c

k e f

l h i

detA, y =

a j c

d k f

g l i

detA, z =

a b j

d e k

g 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 0

3 8 3

0 2 1

= (8 + 0 + 0) − (0 + 6 +12) = −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 0

92 8 3

18 2 1

−10=

−120

−10=12

H =

1 16 0

3 92 3

0 18 1

−10=

−10

−10=1

O =

1 4 16

3 8 92

0 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 0

0 1

⎡

⎣ ⎢

⎤

⎦ ⎥

€

I =

1 0 0

0 1 0

0 0 1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

INVERSE MATRIX

The inverse of the matrix

€

A =a b

c d

⎡

⎣ ⎢

⎤

⎦ ⎥

is

A−1 =1

A

d −b

−c a

⎡

⎣ ⎢

⎤

⎦ ⎥

A−1 =1

ad − cb

d −b

−c a

⎡

⎣ ⎢

⎤

⎦ ⎥

provided

ad − cb ≠ 0

EXAMPLE

Find the inverse of

€

A =3 1

4 2

⎡

⎣ ⎢

⎤

⎦ ⎥

Solution:

€

A−1 =1

6 − 4

2 −1

−4 3

⎡

⎣ ⎢

⎤

⎦ ⎥=

1

2

2 −1

−4 3

⎡

⎣ ⎢

⎤

⎦ ⎥=

1−1

2

−23

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

CHECK THE SOLUTION

€

Show

AA−1 = I = A−1A

3 1

4 2

⎡

⎣ ⎢

⎤

⎦ ⎥1 −

1

2

−23

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥=

1 0

0 1

⎡

⎣ ⎢

⎤

⎦ ⎥,

and

1 −1

2

−23

2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

3 1

4 2

⎡

⎣ ⎢

⎤

⎦ ⎥=

1 0

0 1

⎡

⎣ ⎢

⎤

⎦ ⎥

SOLVING SYSTEMS USING INVERSE MATRICES

SOLVING A LINEAR SYSTEM

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

Writing the original matrix equation.

€

−3 4

2 −1

⎡

⎣ ⎢

⎤

⎦ ⎥x

y

⎡

⎣ ⎢

⎤

⎦ ⎥=

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

⎡

⎣ ⎢

⎤

⎦ ⎥=

1

5

4

52

5

3

5

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

X = A−1B =

1

5

4

52

5

3

5

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

5

−10

⎡

⎣ ⎢

⎤

⎦ ⎥=

−7

−4

⎡

⎣ ⎢

⎤

⎦ ⎥=x

y

⎡

⎣ ⎢

⎤

⎦ ⎥

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