Презентация Matrix Equations and Systems of Linear Equations онлайн
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- Тип файла:ppt / pptx (powerpoint)
- Всего слайдов:17 слайдов
- Для класса:1,2,3,4,5,6,7,8,9,10,11
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Слайды и текст к этой презентации:
№1 слайд
Содержание слайда: Learning Objectives for Section 4.6
The student will be able to formulate matrix equations.
The student will be able to use matrix equations to solve linear systems.
The student will be able to solve applications using matrix equations.
№2 слайд
Содержание слайда: Matrix Equations
Let’s review one property of solving equations involving real numbers. Recall
If ax = b then x = , or
A similar property of matrices will be used to solve systems of linear equations.
Many of the basic properties of matrices are similar to the properties of real numbers, with the exception that matrix multiplication is not commutative.
№3 слайд
Содержание слайда: Basic Properties of Matrices
Assuming that all products and sums are defined for the indicated matrices A, B, C, I, and 0, we have
Addition Properties
Associative: (A + B) + C = A + (B+ C)
Commutative: A + B = B + A
Additive Identity: A + 0 = 0 + A = A
Additive Inverse: A + (-A) = (-A) + A = 0
№4 слайд
Содержание слайда: Basic Properties of Matrices
(continued)
Multiplication Properties
Associative Property: A(BC) = (AB)C
Multiplicative identity: AI = IA = A
Multiplicative inverse: If A is a square matrix and A-1 exists, then AA-1 = A-1A = I
Combined Properties
Left distributive: A(B + C) = AB + AC
Right distributive: (B + C)A = BA + CA
№5 слайд
Содержание слайда: Basic Properties of Matrices
(continued)
Equality
Addition: If A = B, then A + C = B + C
Left multiplication: If A = B, then CA = CB
Right multiplication: If A = B, then AC = BC
№6 слайд
Содержание слайда: Solving a Matrix Equation
Reasons for each step:
Given; since A is n x n,
X must by n x p.
Multiply on the left by A-1.
Associative property of matrices
Property of matrix inverses.
Property of the identity matrix
Solution. Note A-1 is on the left of B. The order cannot be reversed because matrix multiplication is not commutative.
№7 слайд
Содержание слайда: Example
Example: Use matrix inverses to solve the system
№8 слайд
Содержание слайда: Example
Example: Use matrix inverses to solve the system
Solution:
Write out the matrix of coefficients A, the matrix X containing the variables x, y, and z, and the column matrix B containing the numbers on the right hand side of the equal sign.
№9 слайд
Содержание слайда: Example
(continued)
Form the matrix equation AX = B. Multiply the 3 x 3 matrix A by the 3 x 1 matrix X to verify that this multiplication produces the 3 x 3 system at the bottom:
№10 слайд
Содержание слайда: Example
(continued)
If the matrix A-1 exists, then the solution is determined by multiplying A-1 by the matrix B. Since A-1 is 3 x 3 and B is 3 x 1, the resulting product will have dimensions 3 x 1 and will store the values of x, y and z.
A-1 can be determined by the methods of a previous section or by using a computer or calculator. The resulting equation is shown at the right:
№11 слайд
Содержание слайда: Example
Solution
The product of A-1 and B is
№12 слайд
Содержание слайда: Another Example
Example: Solve the system on the right using the inverse matrix method.
№13 слайд
Содержание слайда: Another Example
Example: Solve the system on the right using the inverse matrix method.
Solution:
The coefficient matrix A is displayed at the right. The inverse of A does not exist. (We can determine this by using a calculator.) We cannot use the inverse matrix method. Whenever the inverse of a matrix does not exist, we say that the matrix is singular.
№14 слайд
Содержание слайда: Cases When Matrix Techniques
Do Not Work
There are two cases when inverse methods will not work:
1. If the coefficient matrix is singular
2. If the number of variables is not the same as the number of equations.
№15 слайд
Содержание слайда: Application
Production scheduling: Labor and material costs for manufacturing two guitar models are given in the table below: Suppose that in a given week $1800 is used for labor and $1200 used for materials. How many of each model should be produced to use exactly each of these allocations?
№16 слайд
Содержание слайда: Solution
Let x be the number of model A guitars to produce and y represent the number of model B guitars. Then, multiplying the labor costs for each guitar by the number of guitars produced, we have
30x + 40y = 1800
Since the material costs are $20 and $30 for models A and B respectively, we have
20x + 30y = 1200.
№17 слайд
Содержание слайда: Solution
(continued)
The inverse of matrix A is
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