Matrices - Prince of Songkla University
Matrices
Consider the following problem:
The Acrosonic Company manufactures 4 different loundspeaker systems in 3 separate locations. The output for the month of May is summarized in the following table:
| |Model A |Model B |Model C |Model D |
|Location I |320 |280 |460 |280 |
|Location II |480 |360 |580 |0 |
|Location III |540 |420 |200 |880 |
If we agree to preserve the location of each of the entries in the above table, the set of data may be summarized further as
|320 |280 |460 |280 |
|480 |360 |580 |0 |
|540 |420 |200 |880 |
A is called a Matrix.
A matrix is an ordered rectangular array of real numbers of the form
|a11 |A12 |… |a1n |
|a21 |A22 |… |a2n |
|… | | |… |
|am1 |Am2 |… |amn |
Where m, n are positive integers.
The real numbers aij ,m, i =1, 2, … , m; j = 1, 2, … , n, comprising the array are called the entries or elements of the matrix.
aij is the entry located at the ith row and jth column.
The entries comprising a row in the array are referred to as a row of the matrix.
The entries comprising a column in the array are referred to as a column of the matrix.
Example
A =
B =
C =
D = [ 7 4 -3 1 5 7 ]
A matrix having m rows and n columns is said to have order or size of m x n.
Examples
Matrix A has order 2 x 3
Matrix B has order 3 x 2
Matrix C has order 4 x 1
Matrix D has order 1 x 6
A matrix of order 1 x n is referred to as a row matrix or row vector (of dimension n).
A matrix of order m x 1 is referred to as a column matrix or column vector (of dimension m).
A matrix having the same number of rows and columns is called a square matrix.
Examples
1.
|-3 |1 |4 |
|7 |2 |3 |
|2 |4 |5 |
is a square matrix of order 3 x 3 or simply of order 3.
2.
|4 |7 |3 |9 |8 |
|2 |2 |2 |4 |9 |
|1 |3 |1 |1 |7 |
|4 |8 |1 |7 |3 |
|5 |4 |3 |7 |2 |
is a square matrix of order 5.
Two matrices are said to be equal if they have the same order and their corresponding entries are equal.
Example
|2 |3 |1 | |(3-1) |3 |1 |
|4 |6 |2 | |4 |(4+2) |2 |
|1 |3 |5 | | 1 |2 |
|2 |4 |3 | |3 |4 |
| | | | |5 |3 |
since the matrix on the left has order 2*3 while the matrix on the right has order 3*2, and
|2 |3 | | 2 |3 |
|4 |6 | |4 |7 |
Two matrices A and B of the same order can be added (subtracted) to produce a matrix of the same order.
Examples
1.
|1 |3 |4 | |1 |4 |3 |
|-1 |2 |0 | |6 |1 |-2 |
|1 + 1 |3 + 4 |4 + 3 |
|-1 + 6 |2 + 1 |0 + (-2) |
|2 |7 |7 |
|5 |3 |-2 |
2.
|1 |2 | |2 |-1 |
|-1 |3 | |3 |2 |
|4 |0 | |-1 |0 |
|1 – 2 |2 - (-1) | |-1 |3 |
|-1 - 3 |3 – 2 | |-4 |1 |
|4 – (-1) |0 – 0 | |5 |0 |
Consider the matrix
|320 |280 |460 |280 |
|480 |360 |580 |0 |
|540 |420 |200 |880 |
represents the output of loundspeaker systems of Acrosonic Company.
a) What is the order of the matrix P?
b) Find a24 and give an interpretation of this number.
c) Find the sum of the entries comprising the first row of P and interpret the result.
d) Find the sum of the entries comprising the fourth column of P and interpret the result.
Example
The total output of the Acrosonic Company for the month of June is summarized in the table.
| |Model A |Model B |Model C |Model D |
|Location I |210 |180 |330 |180 |
|Location II |400 |300 |450 |40 |
|Location III |420 |280 |180 |740 |
Find the total output of the company for May and June.
Solution
Let A and B be the matrices that represent the production of the company for the month of May and June respectively.
|320 |280 |460 |280 |
|480 |360 |580 |0 |
|540 |420 |200 |880 |
|210 |180 |330 |180 |
|400 |300 |450 |40 |
|420 |280 |180 |740 |
|320 + |280 + |460 + |280 + |
|210 . |180 . |330 . |180 . |
|480 + |360 + |580 + | 0 + |
|400 . |300 . |450 . |40 . |
|540 + |420 + |200 + |880 + |
|420 . |280 . |180 . |740 . |
|530 |460 |790 |460 |
|880 |660 |1030 |40 |
|960 |700 |380 |1620 |
The following laws hold for matrix addition.
If A, B and C are matrices of the same order, then
1. A + B = B + A (Commutative law)
2. (A + B) + C = A + (B + C)
= A + B + C (Association law)
A zero matrix is a matrix with all its entries are zero.
If O is a zero matrix having the same order as a non-zero matrix A, then
A + O = O + A = A
A matrix B may be multiplied by a real number, called a scalar in the context of matrix algebra. The scalar product, denoted by cB, is a matrix obtained by multiplying each entry of B by c.
Example
The scalar product of the matrix
|3 |-1 |2 |
|0 |1 |4 |
By the scalar 3 is the matrix
|3 |-1 |2 | |9 |-3 |6 |
|0 |1 |4 | |0 |3 |12 |
Example
The management of the Acrosonic company has decided to increase its July production of loundspeaker systems by 10% (over its June output). Find a matrix giving the targeted production for the month of July.
Solution
The matrix represents the total output of the company for the month of June is
|210 |180 |330 |180 |
|400 |300 |450 |40 |
|420 |280 |180 |740 |
The required matrix is given by
|210 |180 |330 |180 |
|400 |300 |450 |40 |
|420 |280 |180 |740 |
|231 |198 |363 |198 |
|440 |330 |495 |44 |
|462 |308 |198 |814 |
Example Given
A = 3 4 and B = 3 2
-1 2 -1 2
Find the matrix X satisfying the matrix equation 2X + B = 3A.
Solution
From 2X + B = 3A
2X = 3A –B
= 3 3 4 - 3 2
-1 2 -1 2
= 9 12 - 3 2
-3 6 -1 2
2X = 6 10
-2 4
X = 1 6 10 = 3 5
2 -2 4 -1 2
If A is an m x n Matrix with elements aij , i = 1, …, m, j = 1, …,n, then the transpose of A, denoted by AT, is the n x m matrix with elements aji, j = 1, …, n, i = 1, …,m. Thus the matrix AT is obtained by interchanging the rows and columns of A.
Examples
A = 1 2 3 Then AT = 1 4
4 5 6 2 5
3 6
Exercises
1. Let
A = 2 -3 9 -4
-11 2 6 7
6 0 2 9
5 1 5 -8
a) What is the order of A?
b) Find a14, a21, a31 and a43.
2. Perform
1 1 0 0 -4 + 4 3 0 –1 4 - 1 3 -9 -1 0
2 3 0 –1 6 3 -2 1 -6 2 3 6 2 0 -6
-2 1 -4 2 8 2 0 –2 0 1 -3 1
3. Solve for u, x, t and z for the matrix equations:
2X –1 3 2 3 U 2
2 4 Y-2 = 2 4 5
2Z -3 2 4 -3 2
4. Verify by direct computation (3 + 5)A = 3A + 5A
Where
3 1
A = 2 4
-4 0
5. Find the transpose of the matrix
1 2 4 6
2 3 2 5
6 2 3 0
4 5 0 2
Consider the following problem:
On a certain day AI’s Service Station sold 800 gallons of premium, 1200 gallons of regular, 1000 gallons of super-unleaded, and 1600 gallons of regular-unleaded gasoline. Suppose the price of gasoline on this day was $1.80/gallons for the premium, $1.30/gallons for regular, $1.60/gallons for super-unleaded, and $1.50/gallons for regular-unleaded gasoline. Find the total revenue realized by AI’s Service Station for that day.
Generally, if A is a matrix of order m x n and B is a matrix of order n x p (so that the number of columns of A is equal to the number of rows of B) then the matrix product of A and B, AB, is defined and is a matrix of order m x p.
Schematically,
Order of A Order of B
m x n n x p
Order of AB
m x p
The day’s sale of gasoline may be represented by the 1 x 4 matrix
A = [ 800 1200 1000 1600 ]
And let the unit selling price of premium, regular, super-leaded and regular-leaded gasoline be the entries in the 4 x 1 matrix
1.80
1.30
B = 1.60
1.50
AB = [ 800 1200 1000 1600 ] 1.80
1.30
1.60
1.50
= [ (800)(1.8) + (1200)(1.3) + (1000)(1.6) + (1600)(1.5) ]
= [ 7000 ]
Example
a11 a12 a13 b11 b12 b13 b14
A = a21 a22 a23 B = b21 b22 b23 b24
b31 b32 b33 b34
Order 2 x 3 order 3 x 4
C = AB order 2 x 4
= C11 C12 C13 C14
C21 C22 C23 C24
C11 = [ a11 a12 a13 ] b11
b21
b31
= [ c11b11 + a12b21 + a13b31 ]
C12 = [ a11 a12 a13 ] b12
b22
b32
= [ a11b12 + a12b22 + a13b32 ]
The order entries in C computed in a similar manner.
C21 = [ -1 2 3 ] 1 = -1 + 8 +6 = 13
4
2
C22 = [ -1 2 3 ] 3 = -3 - 2 + 12 = 7
-1
4
C23 = [ -1 2 3 ] -3 = 3 + 4 +3 = 10
2
1
C = AB = 15 24 -3
13 7 10
Example
Let A = 3 1 4 and B = 1 3 -3 compute AB.
-1 2 3 4 -1 2
2 4 1
Solution
Let C = AB
Order of A = 2 x 3
Order of B = 3 x 3
(Order of C = 2 x 3
C = C11 C12 C13
C21 C22 C23
C11 = [ 3 1 4 ] 1 = 3 + 4 + 8 = 15
4
2
C12 = [ 3 1 4 ] 3 = 9 + (-1) + 16 = 24
-1
4
C13 = [ 3 1 4 ] -3 = -9 + 2 + 4 = -3
2
1
Example
If A = 3 2 1 , B = 1 3 4
-1 2 3 2 4 1
3 1 4 -1 2 3
Then
6 19 17
AB = 0 11 7
1 21 25
12 12 26
BA = 5 13 18
4 5 17
In general AB BA for any two square matrices A and B. However, the following properties are valid for matrix multiplication.
A(BC) = (AB)C (Associate law)
A(B + C) = AB + AC (Distributive law)
Whenever the products are defined.
Example
The Ace Novelty Company received an order from the Magic World Amusement Park for 900 “Giant Pandas”, 1200 “Saint Bernards” and 200 “Big Birds”.
Management decided that 500 of Giant Pandas, 800 of Saint Bernards and 1300 of Big Birds could be manufactured in their Los Angeles plant with the balance of the order to be filled from their Seattle plant. Each Giant Panda requires 1.5 square yards of plush, 30 cubic feet of stuffing material, and 5 pieces of trimming, each Saint Bernard requires 2 square yards of plush, 3.5 cubic feet of stuffing material, and 8 pieces of trimming; and each Big Bird requires 2.5 square yards of plush, 25 cubic feet of stuffing material, and 15 pieces of trimming. The plush costs $4.5 per square yard, the stuffing material costs $.10 per cubic foot, and the trimming costs $.25 per unit.
a) Find the amount of each type of material to be purchase for each of the two plants.
b) What is the total cost of materials incurred by each plant and the total cost of materials incurred by Ace Novelty Company in fulfilling the order?
Solution
The quantities of each type of stuffed animal to be produced at each of the two plant locations may be expresses as 2 x 3 production matrix P:
G. Panda S.Bernards B.Birds
P = L.A. 500 800 1300
Seattle 400 400 700
The amount and type of material required to manufactured each type of stuffed animal may be expressed as 3 x 3 activity matrix A:
Plush S. Mat. Trim
G. Pandas 1.5 30 5
A = S. Bernards 2 3 8
B. Birds 2.5 25 15
The unit cost for each type of material may be expressed as 3 x 1 matrix C:
Plush 4.5
C = S. Mat. 0.10
Trim 0.25
a) The amount of each type of material to be required for each of the two plants is given by the matrix PA:
1.5 30 5
PA = [ 500 800 1300 ] 2 35 8
2.5 25 15
Plush S. Mat. Trim
= L.A. 5600 75500 28400
Seattle 3150 43500 15700
b) The total cost of materials to be incurred by each plant is given by the matrix PAC:
4.5
PAC = 5600 75500 28400 0.10 = L.A. 39850
3150 43500 15700 0.25 Seattle 22450
The total cost of materials to be incurred by the company is $39850 + $22450 = $62300
Exercise
The following figure shows the routes of an international airline connecting five cities. A line joining two cities indicates that there is a direct flight between them.
Tokyo Los Angleles
Hong Kong Taipai
Thailand
The same information can be represented by the following matrix A.
Thailand H.K. Taipai Tokyo L.A.
Thailand 0 1 1 0 0
H.K. 1 0 1 1 0
A = Taipai 1 1 0 1 1
Tokyo 0 1 1 0 1
L.A. 0 0 1 1 0
Where aij = 1 indicates that there is a direct flight between the ith city and the jth city, and aij = 0 indicates that there is no direct flight between the cities.
a) Compute A2
b) By referring to the figure, verify that the entry, aij of A2 represents the number of one-stop routes between the ith city and the jth city.
Exercise
The following figure shows the routes of a domestic airline connecting five cities:
Boston
Madison
Chicago New York City
Philadelphia
a) Represent this information by a matrix A.
b) Write the matrix that shows the number of one-stop routes between any two cities.
A square matrix of order n having ones along the main diagonal and zeroes elsewhere is called an identity matrix of order n. Thus, if In denotes the identity matrix of order n, then
1 0 0 … 0
0 1 0 … 0
In = …
0 0 0 … 1
The identity matrix has the property that InA = A for any n x r matrix A, and BIn = B for any s x n matrix B. In particular, if A is a square matrix of order n then InA = AIn = A.
Inverse of a matrix
Let A be a matrix of dimension n x n.
A matrix B of dimension n x n is called the inverse of the matrix A (denoted by A-1), if
AB = BA = I
AA-1 = A-1A = I
Example
Show that ½ -½ is the inverse of 2 1
0 1 0 1
Solution A A-1 I
2 1 ½ -½ = 1 0
0 1 0 1 0 1
and ½ -½ 2 1 = 1 0
0 1 0 1 0 1
( ½ -½ is the inverse of 2 1
0 1 0 1
Example
Find the inverse of A = 2 1
0 1
Solution
Assume that A-1 = a b
c d
A.A-1 = 2 1 a b = 1 0
0 1 c d 0 1
( 2a+c 2b+d = 1 0
c d 0 1
( 2a + c = 1
2b + d = 0
c = 0
d = 1
Thus a = ½ , c = 0
b = -½ , d = 1
Hence
A-1 = ½ -½
0 1
A nonsquare matrix has no inverse.
Since whenever a matrix A has an inverse B, then
AB = BA = I
But A is not a square matrix which causes AB and BA to have different dimensions and prevents them from being equal. So, such a B could not exist.
Sometimes, a square matrix does not have an inverse.
Example
Show that the matrix A = 0 1 does not have an inverse.
0 0
Solution
Assume that A-1 = a b
c d
A.A-1 = I
Thus 0 1 a b = 1 0
0 0 c d 0 1
c d = 1 0
0 0 0 1
These two matrices can never be equal (since 0 1) therefore the assumption that A has an inverse is false.
-----------------------
A =
3 4 2
2 2 1
3. -2
1 1
3 0
4
3
2
1
is the matrix of 4 rows and 1 column
is a matrix of 2 rows and 3 columns
is the matrix of 1 row and 6 columns
is the matrix of 3 rows and 2 columns
A =
B =
=
(
(
+
=
-
=
=
=
P =
P =
A =
( A =
P =
B =
=
A + B =
3A = 3
=
B =
P =
=
1.1B = 1.1
=
=
but
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