🗊Презентация Matrix Equations and Systems of Linear Equations

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Matrix Equations and Systems of Linear Equations, слайд №1Matrix Equations and Systems of Linear Equations, слайд №2Matrix Equations and Systems of Linear Equations, слайд №3Matrix Equations and Systems of Linear Equations, слайд №4Matrix Equations and Systems of Linear Equations, слайд №5Matrix Equations and Systems of Linear Equations, слайд №6Matrix Equations and Systems of Linear Equations, слайд №7Matrix Equations and Systems of Linear Equations, слайд №8Matrix Equations and Systems of Linear Equations, слайд №9Matrix Equations and Systems of Linear Equations, слайд №10Matrix Equations and Systems of Linear Equations, слайд №11Matrix Equations and Systems of Linear Equations, слайд №12Matrix Equations and Systems of Linear Equations, слайд №13Matrix Equations and Systems of Linear Equations, слайд №14Matrix Equations and Systems of Linear Equations, слайд №15Matrix Equations and Systems of Linear Equations, слайд №16Matrix Equations and Systems of Linear Equations, слайд №17
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Слайд 1


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.
Описание 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.

Слайд 2


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.
Описание 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.

Слайд 3


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
Описание 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

Слайд 4


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
Описание 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

Слайд 5


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
Описание 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

Слайд 6


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.
Описание 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.

Слайд 7


Example Example: Use matrix inverses to solve the system
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Example Example: Use matrix inverses to solve the system

Слайд 8


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.
Описание 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.

Слайд 9


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:
Описание 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:

Слайд 10


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:
Описание 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:

Слайд 11


Example Solution The product of A-1 and B is
Описание слайда:
Example Solution The product of A-1 and B is

Слайд 12


Another Example Example: Solve the system on the right using the inverse matrix method.
Описание слайда:
Another Example Example: Solve the system on the right using the inverse matrix method.

Слайд 13


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.
Описание 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.

Слайд 14


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.
Описание 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.

Слайд 15


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?
Описание 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?

Слайд 16


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.
Описание 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.

Слайд 17


Solution (continued) The inverse of matrix A is
Описание слайда:
Solution (continued) The inverse of matrix A is

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