MATRICES &
DETERMINANTS
Monika V Sikand
Light and Life Laboratory
Department of Physics and Engineering physics
Stevens Institute of Technology
Hoboken, New Jersey, 07030.
OUTLINE
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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 rows
2 0 5 
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
2 4


1 7
A matrix with all zero
entries
0 0


0 0
3
2
 
3

Zero matrix

2 1 4
MATRIX OPERATIONS
COMPARING MATRICES
EQUAL MATRICES: Matrices having equal corresponding
entries.
For Example:
 5
 4


 4
0  5
0 
3  

1 0.75



4
2 6  2 6 

 

0 3  3 2
ADDING MATRICES
Matrices of same dimension can be added
For Example:
 3  1  31  4 
    
  
4 0 4  0 4

2 
 
2
 
2  2 
 
5 



SUBTRACTING MATRICES
Matrices of same dimension can be subtracted
For example:
8 3 2 7 8  2 3 (7) 6 10

 
 
 

4 0 6 1 4  6 0  (1) 2 1 
MULTIPLYING A MATRIX BY A
SCALAR
For example:
1

20

4
2

 0

8
6

 6

6
2 4 5   (2)1 (2)  2 4 5 
 
 
 

3  6 8  (2)0
(2)3  6 8
5 
 
2 6 
 
(2)  4 (2)5 
 
2 6 

4  4 5 
 

6  6 8
10
 
2 6 

9 

14
4 

SOLVING A MATRIX
EQUATION
For example:
Solve:
3x 1 4 1  26 0
2
 
 

8
5
2
y
12
8
 
 


3x  4 11 26 0
2
 

8  2 5  y  12 8
6x  8
0  26 0

 

12
10

2y
12
8

 

Equate:
6x  8  26
x3
10  2y  8
y 1
MULTIPLYING MATRICES
PRODUCT OF TWO MATRICES
For example:
3
A  
1
1
B  
2
2

0
4

1 
FIND (a.) AB and (b.) BA
SOLUTION
3 21 4
AB  


1 02 1
7 10
AB  

1 4 
1 43 2
BA  


2
1
1
0



7 2
BA  

5
4


SIMPLIFY
2 1
2 0
1 1
A  
, B  
, C  

1 3
4 2
3 2
Simplify:
a.) A(B+C)
b.) AB+AC
SOLUTION
A(B+C):
2 12
 

1 31
2 11
 

1 37
5 6 
 

22
11


1 1 1
 

3 3 2
1

4
SOLUTION
AB+AC:
2
 
1
0
 
14
5
 
22
12

34
2 5
 
6 8
6 

11
0 2 11 1
 


2 1 33 2
4

5
DETERMINANTS &
CRAMER”S RULE
DETERMINANT OF 22 MATRIX
The determinant of a 22 matrix is the difference of
the entries on the diagonal.
a b
det
 ad  bc
c d 
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 33 MATRIX
The determinant of a 33 matrix is the difference in the
sum of the products in red from the sum of the products
in black.
a b

detd e

g h
c  a b

f  d e
i 
 g h
ca b
fd e
i 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

Solution:
2 1 3 2 1
2 0 1 2 0
1
2
4 1
2
 [0  (1)  (12)] (0  4  8)
 1312
 25
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.
N
Bermuda (938,454)
W
.
.
Miami (0,0)
.
S
E
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:
938 454 1
1
Area   900 518 1
2
0
0
1
1
Area   [(458, 884 0  0)  (0  0  408, 600)]
2
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.
N
New Delhi (100,120)
.
W
.
Jaipur (0,0)
.
S
E
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:
100 120 1
1
Area   140
2
0
20
0
1
1
1
Area   [(2000 0  0)  (0  0 16800)]
2
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.
N
(35,220)
.
.
W
(0,0)
S
.(112,56)
E
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:
35
1
Area   112
2
0
220 1
56
0
1
1
1
Area   [(1960 0  0)  (0  0  24640)]
2
Area  11340
Hence, area of the region is about 11340 square miles.
CRAMER”S RULE FOR A 22
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:
e
b
f d
x
det A
a e
c f
y
det A
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.
2
5
10 4 8  (50) 42
x


 1
42
42
42
8 2
2 10 80  4 84
y


2
42
42
42
The solution is (-1,2)
CRAMER”S RULE FOR A 33
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:
j b c
a j c
a b j
k e f
d k f
d e k
l h i
g l i
g h l
x
,y 
,z 
det A
det A
det A
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
Write a linear system using the formula for each compound
C + 4H
= 16
3C+ 8H + 3O = 92
2H + O =18
Evaluate the determinant of the coefficient matrix.
1 4 0
3 8 3  (8  0  0)  (0  6 12)  10
0 2 1
SOLUTION
Apply cramer’s rule since determinant is not zero.
16 4 0
92 8 3
18 2 1 120
C

 12
10
10
1 16 0
3 92 3
0 18 1 10
H

1
10
10
1 4 16
3 8 92
0 2 18 160
O

 16
10
10
Atomic weight of carbon = 12
Atomic weight of hydrogen =1
Atomic weight of oxygen =16
IDENTITY AND INVERSE
MATRICES
IDENTITY MATIX
22 IDENTITY MATRIX
1 0
I  

0 1
33 IDENTITY MATRIX
1 0 0


I  0 1 0

0 0 1

INVERSE MATRIX
The inverse of the matrix
a b 
A  

c
d


is
1 d b
A  

A c a 
1 d b
1
A 


ad  cb c a 
provided
1
ad  cb  0
EXAMPLE
Find the inverse of
3 1
A  

4 2
Solution:


1





2
1
2
1
1
1
A1 

 
 
6  4 4 3  2 4 3  2

1
2 
3 

2 
CHECK THE SOLUTION
Show
AA1  I  A 1 A

1 
3 11  2  1 0
 
,




4 22 3  0 1

2 
and

1 
1  2 3 1 1 0
 


3 
4
2
0
1
 

2


2 
SOLVING SYSTEMS USING
INVERSE MATRICES
SOLVING A LINEAR SYSTEM
-3x + 4y = 5
2x - y = -10
Writing the original matrix equation.
A
X
B
3 4 x  5 

   
2 1y 10

AX = B
A-1AX = A-1B
IX = A-1B
X = A-1B
USING INVERSE MATRIX TO
SOLVE THE LINEAR SYSTEM
-3x + 4y = 5
2x - y = -10
1 4 
1 1 4 5 5 
1
A 

 2 3 
3  8 2 3 

5 5 
1 4 
5 5  5  7 x
1
X  A B  
    
2 3 

10 4 y
5 5 
Hence the solution of the system is (-7,-4)