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)