Title: LESSON 2 MATRICES
1LESSON 2 MATRICES
2Basics of Matrices
- A matrix is a rectangular array of ordered
numbers. - Example. Let A denote the matrix
- This matrix A has three rows and four columns and
is said to be a 3 x 4 matrix or of dimension 3x4.
- We denote the element on the second row and
fourth column with a2,4.
3Solving a System Using Augmented Matrices
(GAUSS-JORDAN)
- Equation 1 2x 3y 2
- Equation 2 5x 4y 12
Note equation system can be written in matrix
notation as above. Coefficient matrix times
variable column vector equals vector of
constants. Also note conformability is satisfied
(2x2) (2x1) (2x1).
4Augmented Matrix
- Augmented matrix corresponding to the system
- Our goal is to use row operations to transform
our augmented matrix into an augmented matrix of
the form
5Reduced Row Echelon Form
- For an m x n matrix to be in reduced row echelon
form, it must satisfy the following - 1. All rows containing only zeros are at the
bottom of the matrix - 2. The first (leftmost) nonzero entry in a
nonzero row is a 1. This is called the leading 1
of its row. - 3. The leading 1 of each nonzero row lies left of
the leading 1 of any lower row - 4. For any column that has a leading 1, the other
entries in the column are zeros
6Reduce Echelon Form
- Elementary Operations
- 1 Interchange any two rows
- 2 Multiply any row by any non zero constant
- 3 Add any multiple of any row to any other row.
7Steps
- First step get a 1 in the upper left by using
R1only. - Second get a 0 in the lower left by using R1R2
to get R2. - Third get a 1 in the 2nd row, 2nd column by using
R2 only. - Fourth get a 0 in the 1st row and 2nd column by
using R2R1to get R1 - Multiply R1 by ½
8Continued
- Solution
- Step 2 multiply R1 by 5 and add R2 to get new R2
9Continued
- Step 3 multiply R2 by 2/7
- Step 4 multiply R2 by 3/2 and add R1 to get R1
10Example Solving a System Using Augmented Matrices
- 1. Structure the following system in matrix
notation. - Equation1 x y 2z 20
- Equation2 x 2y 30
- Equation3 x y z 20
11Continued
12Continued
- Multiply R1 by (-). Add R1to R2 and R1 to R3
13Continued
- Multiply R3 by (-). MultiplyR3 by 2 and add to
R2. Multiply R3 by 2 and Add to R1.
14Continued
- Multiply R2 by (-). Add R2 to R1.
15MATRICES OPERATIONS
16Sum or Difference of Matrices
- To add(subtract) two matrices they must be of the
same order. This means that they must have the
same number of rows and columns. - To add(subtract) two such matrices, we simply
add(subtract) the corresponding elements.
17Sum Difference of Matrices
18 Matrix
- To multiply a matrix by a scalar means to
multiply that matrix by a single number. - In order to perform scalar multiplication, we
merely multiply each element in the matrix by
this number(ie. the scalar).
19Example of Scalar Multiplication
20TRANSPOSE
- If A is a m x n matrix with entries ai,k, then
the transpose of A denoted A is a n x m matrix
with entries ak,i,
T
21Matrix Multiplication
- In this case we want to develop the technique for
multiplying two matrices. - This product is defined only if matrix A is (m x
n) and matrix B is a (n x p).So the number of
columns in A has to be equal to the number of
rows in B. Matrices are said to be
conformable.The product, C AB then is a (m x
p) matrix.The element of the ith row and the jth
column of the product is found by multiplying the
ith row of A by the jth column of B. ci,j sumk
(ai,k.bk,j)
22Example of Matrix Multiplication
Here the of columns in A is equal to the of
rows in B
23Example of Matrix Multiplication
Find (a) AB and (b) BA
To obtain the entries in the first row of AB,
multiply the first row 1,3 of A by the columns
of B
24Example of Matrix Multiplication
THUS AB
25Example of Matrix Multiplication
- B is a 2x3 and A is a 2x2. And the number of
columns in B is not equal to the number of rows
in A. Hence, the product BA is not defined.
26MATRIX INVERSE AND SOLVING SYSTEMS OF LINEAR
EQUATIONS
27Identity Matrix
- An Identity matrix is a matrix that behaves like
the multiplicative identity 1. If I is an
identity matrix and A is another matrix then
IAA. An identity matrix exists only for square
matrices. The identity matrix of dimension m x n
has 1s along the diagonal from upper to lower
right and 0s elsewhere.
28Continued
- A square matrix is a matrix with the same number
of rows as columns. - An n x n square matrix is said to be of order n
and is sometimes called an s-square matrix. The
operations of addition, multiplication, scalar
multiplication, and transpose can be performed on
any n-square matrices, and the result is again an
n-square matrix.
29Invertible Matrices
- A square matrix A is said to be invertible if
there exists a matrix B with the property that - ABBAI, the identity matrix
- Such a matrix B is unique and it is called the
inverse of A and is denoted by - Note that A is the inverse of B if and only if B
is the inverse of A.
30Identity
31Inverses
- Example A x BI
- Thus A and B are inverses
32Find the Inverse of a Matrix
- To find the inverse of Matrix A
- The inverse should have a form
- If we multiply A x Inverse
33Continued
- This equation is the same as the systems of
linear equations - 1a2c1 and 2a3c0
- 1b2d0 and 2b3d1
- From previous examples (SLIDE 4 Solving a System
Using - Augmented Matrices (GAUSS-JORDAN)) we write an
- augmented matrix
34Continued
- First step get a 1 in the upper left by using
R1only. - Second get a 0 in the lower left by using R1R2
to get R2. - Third get a 1 in the 2nd row, 2nd column by using
R2 only. - Fourth get a 0 in the 1st row and 2nd column by
using R2R1to get R1
35Continued
- Solution multiply R1 by 2 and add to R2
- Multiply R2 by 2 and add to R1
36Continued
- Multiply R2 by 1
- Solution
37Example
- Solve the system of linear equations by using the
inverse of the coefficient matrix. - x 3y 7
- 4x 2y 9
- Write the augmented matrix
38Continued
- First step get a 1 in the upper left by using
R1only. - Second get a 0 in the lower left by using R1R2
to get R2. - Third get a 1 in the 2nd row, 2nd column by using
R2 only. - Fourth get a 0 in the 1st row and 2nd column by
using R2R1to get R1 - Solution we have a 1 in the upper left.
39Continued
- Multiply R2 by 1/4 and add to R1
- Multiply R2 by 2/5
40Continued
- Multiply R2 -3 and add to R1
41Using a matrix inverse to Solve A System
bConstants
ACoefficients
XVariables
So equation system is Ax b
42Continued
43Example
- 1. Structure the following system in matrix
notation. - Equation1 x y 2z 20
- Equation2 x 2y 30
- Equation3 x y z 20
44Finding an Inverse Matrix
- Finding an Inverse Matrix
- Make the augmented matrixA I
- Use the row operations to reduce A I to the
form I A - Write the augmented Matrix
-1
45Continued
- Use the Gauss Jordan Method to solve.
- Multiply R1 by-1 and add R2 and R3
- Multiply R2 by 1 and add R1
46Continued
- Multiply R3 by 2 and add R2 and multiply R3 by 4
and add R1 - Multiply R3 by -1
47Solve Previous Equation System
A inverse
48Using a matrix inverse to Solve A System
bConstants
ACoefficients
XVariables
So equation system is Ax b
49Solve Previous Equation System
A inverse
50Continued
- If you multiply A inverse x the constants we will
find the solution
51Solving
A inverse
b
x
52Solving
Note this provides the solutions for x, y, z.
Thus, x10, y10, 7 z0.
53Leontief Input-Output Models
- Wassily Leontief created the model.
- He first published it in 1965.
- He received a Nobel Prize in 1973 for this work.
- The basic idea is that the outputs of some
industries are the inputs of others, and you can
keep track of this with a matrix.
54Input-Output Models
- Capture inter-industry transactions
- Industries use the products of other industries
to produce their own products. - For example - automobile producers use steel,
glass, rubber, and plastic products to produce
automobiles. - Outputs from one industry become inputs to
another. - When you buy a car, you affect the demand for
glass, plastic, steel, etc.
55Input-Output Models (page 123)
- Lets consider a simple model of an economy in
which only 3 goods are produced Petroleum,
Transportation, and chemicals. These 3 industries
are called sectors in a 3-sector economy - Any industry that does not produce petroleum,
transportation, or chemicals is said to be
outside the economy. However industries outside
the economy may have demands on the outputs of
these three sectors. These type of demand are
called external demands. - Also, the three sectors tend to use their own
outputs. For example Transportation demand as
input oil from Petroleum. These types of demands
among the three sectors are referred as internal
demands
56Basic Input-Output Internal Demands
Chemicals
Transportation
Petroleum
Petroleum
Transportation
57External Demands
Chemicals
Manufacturing
Agriculture
OUTSIDE THE ECONOMY
58Determining Internal Demand
- Lets consider the following for our three
sectors - The production of 1 worth of petroleum (P)
requires 0.10 of itself, 0.20 of transportation
and 0.40 of chemicals. - The production of 1 worth of transportation (T)
requires 0.10 of itself, 0.60 of petroleum and
0.25 of chemicals. - The production of 1 worth of chemicals (C)
requires 0.20 of itself, 0.20 of petroleum and
0.30 of transportation .
59Continued
- Therefore the input-output matrix T
Output
Petr
Tran
Chem
Petr
Input
Tran
Chem
60Continued
- If the three-sector economy produces 900 million
of P, 850 million of T and 800 million of C,
determine how much of this production is consumed
internally.Lets use a matrix to represent the
information above. -
- Matrix P is the Total Production matrix
61Continued
- Lets determine how much of this production is
consumed internally (also known as internal
demand TxP) . - The internal demand for P, T and C is 760m,
505m and 732 m.
62External Demand
- To determine the external demand matrix D, we
simply subtract internal demand from matrix P. - DP-TP
- Also ProductionInternal Demand External Demand
- PTPD
63Determining Total Output
- Given the technological matrix T and the external
demand matrix D determine the total output matrix
P. - PTPD
- Lets rename P with X
- XTXD or DX-TX or DX(1-T)
- when using a matrix 1 is the identity matrix.
- DX(I-T) now provided that I-T has an inverse we
try to solve for X -
64Continued
XTXD
X-TXD
(I-T)XD
65Continued
Simplification
66Continued
67End Lesson 2