matrices determinants ms
TRANSCRIPT
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MATRICES &DETERMINANTS
Monika V SikandLight and Life Laboratory
Department of Physics and Engineering physicsStevens Institute of TechnologyHoboken, New Jersey, 07030.
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OUTLINE
Matrix Operations
Multiplying MatricesDeterminants and Cramers Rule Identity and Inverse Matrices
Solving systems using Inverse matrices
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MATRIX
A rectangular arrangement of numbers in rows and columns
For example:
6 2 12 0 5
2 rows
3 columns
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TYPES OF MATRICES
NAME DESCRIPTION EXAMPLE
Row matrix A matrix with only 1row
Column matrix A matrix with only Icolumn
Square matrix A matrix with samenumber of rows andcolumns
Zero matrix A matrix with all zeroentries
3 2 1 4 2
3
2 4
1 7
0 0
0 0
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MATRIX OPERATIONS
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COMPARING MATRICES
For Example:
5 044
34
5 0
1 0.75
2 6
0 3
2 6
3 2
EQUAL MATRICES: Matrices having equal correspondingentries.
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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
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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
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MULTIPLYING A MATRIX BY ASCALAR
For example:
2
1 2
0 34 5
4 5
6 82 6
( 2)1 ( 2) 2
( 2)0 ( 2)3( 2) 4 ( 2)5
4 5
6 82 6
2 4
0 6
8 10
4 5
6 8
2 6
6 9
6 14
6 4
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SOLVING A MATRIXEQUATION
For example: Solve :
23 x 1
8 5
4 1
2 y
26 0
12 8
23 x 4 1 1
8 2 5 y
26 0
12 8
6 x 8 0
12 10 2 y
26 0
12 8
Equate :
6 x 8 26
x 3
10 2 y 8
y 1
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MULTIPLYING MATRICES
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PRODUCT OF TWO MATRICES
A 3 21 0
B1 4
2 1
For example:
FIND (a.) AB and (b.) BA
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SOLUTION
AB3 2
1 0
1 4
2 1
AB7 10
1 4
BA1 4
2 1
3 2
1 0
BA7 2
5 4
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SIMPLIFY
A2 1
1 3, B
2 0
4 2,C
1 1
3 2Simplify:a.) A(B+C)
b.) AB+AC
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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):
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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
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DETERMINANTS &CRAMERS RULE
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DETERMINANT OF 2 2 MATRIX
deta b
c d ad bc
The determinant of a 2 2 matrix is the difference of the entries on the diagonal.
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EVALUATE
Find the determinant of the matrix:
1 3
2 5Solution:
1 32 5
1(5) 2(3) 5 6 1
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DETERMINANT OF 3 3 MATRIX
The determinant of a 3 3 matrix is the difference in thesum of the products in red from the sum of the productsin 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)]
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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:
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USING MATRICES IN REAL LIFE
The Bermuda Triangle is a large trianglular region inthe Atlantic ocean. Many ships and airplanes havebeen lost in this region. The triangle is formed by
imaginary lines connecting Bermuda, Puerto Rico, andMiami, Florida. Use a determinant to estimate the areaof the Bermuda Triangle.
EW
N
S
Miami (0,0)
Bermuda (938,454)
Puerto Rico (900,-518)
...
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SOLUTION
The approximate coordinates of the Bermuda Triangles three vertices are: (938,454), (900,-518), and (0,0). Sothe area of the region is as follows:
Area 12
938 454 1
900 518 1
0 0 1
Area 12
[( 458,884 0 0) (0 0 408,600)]
Area 447,242
Hence, area of the Bermuda Triangle is about 447,000square miles.
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USING MATRICES IN REAL LIFE
The Golden Triangle is a large triangular region in theIndia.The Taj Mahal is one of the many wonders that liewithin the boundaries of this triangle. The triangle isformed by the imaginary lines that connect the cities of New Delhi, Jaipur, and Agra. Use a determinant toestimate the area of the Golden Triangle. Thecoordinates given are measured in miles.
EW
N
S
Jaipur (0,0)
New Delhi (100,120)
Agra (140,20)
. ..
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SOLUTION
The approximate coordinates of the Golden Triangles three vertices are: (100,120), (140,20), and (0,0). So thearea of the region is as follows:
Area 12
100 120 1
140 20 1
0 0 1
Area 12
[(2000 0 0) (0 0 16800)]
Area 7400
Hence, area of the Golden Triangle is about 7400 squaremiles.
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USING MATRICES IN REAL LIFE
Black neck stilts are birds that live throughout Floridaand surrounding areas but breed mostly in thetriangular region shown on the map. Use a determinant to estimate the area of this region. The coordinatesgiven are measured in miles.
EW
N
S
(0,0)
(35,220)
(112,56)
. .
.
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SOLUTION
The approximate coordinates of the Golden Triangles three vertices are: (35,220), (112,56), and (0,0). So thearea of the region is as follows:
Area 12
35 220 1
112 56 1
0 0 1
Area 12
[(1960 0 0) (0 0 24640)]
Area 11340
Hence, area of the region is about 11340 square miles.
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CRAMERS RULE FOR A 2 2SYSTEM
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 det A
y
a e
c f
det A
The numerators for x and y are the
determinant of the matrices formed byusing the column of constants asreplacements for the coefficients of x andy, respectively.
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EXAMPLE
Use cramers rule to solve this system:
8x+5y = 22x-4y = -10
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SOLUTION
Solution: Evaluate the determinant of the coefficient matrix
8 5
2 4 32 10 42
Apply cramers rule since the determinant is not zero.
x
2 510 4
428 ( 50)
424242
1
y
8 2
2 1042
80 442
8442 2 The solution is (-1,2)
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CRAMERS RULE FOR A 3 3SYSTEM
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
det A
, y
a j c
d k f
g l i
det A
, z
a b j
d e k
g h l
det A
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EXAMPLE
The atomic weights of three compounds are shown. Use alinear system and Cramers rule to find the atomic weightsof carbon(C ), hydrogen(H), and oxygen(O).
Compound Formula Atomic weight
Methane CH 4 16
Glycerol C 3H8O 3 92
Water H 2O 18
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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 = 16
3C+ 8H + 3O = 922H + O =18
Evaluate the determinant of the coefficient matrix .
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SOLUTION
Apply cramers rule since determinant is not zero.
C
16 4 0
92 8 3
18 2 110
12010
12
H
1 16 0
3 92 3
0 18 110
1010
1
O
1 4 16
3 8 92
0 2 1810
16010
16
Atomic weight of carbon = 12
Atomic weight of hydrogen =1
Atomic weight of oxygen =16
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IDENTITY AND INVERSEMATRICES
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IDENTITY MATIX
2 2 IDENTITY MATRIX 3 3 IDENTITY MATRIX
I 1 0
0 1 I
1 0 0
0 1 0
0 0 1
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INVERSE MATRIX
The inverse of the matrix
A
a b
c d
is
A 11 A
d b
c a
A 11
ad cb
d b
c a
provided
ad cb 0
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EXAMPLE
Find the inverse of
A3 1
4 2
Solution:
A 11
6 4
2 1
4 312
2 1
4 3
11
2
232
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CHECK THE SOLUTION
Show
AA 1 I A 1 A
3 1
4 2
1 12
232
1 0
0 1,
and
112
232
3 1
4 2
1 0
0 1
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SOLVING SYSTEMS USING
INVERSE MATRICES
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SOLVING A LINEAR SYSTEM
-3x + 4y = 52x - y = -10
Writing the original matrix equation.
3 4
2 1
x
y
5
10
A X B AX = B A -1 AX = A -1B
IX = A-1
BX = A -1B
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USING INVERSE MATRIX TOSOLVE THE LINEAR SYSTEM
-3x + 4y = 52x - y = -10
A 11
3 8
1 4
2 3
15
45
25
35
X A 1 B
1
5
4
525
35
510
74
x y
Hence the solution of the system is (-7,-4)