Matrix Equations and Systems of Linear Equations

Содержание

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Barnett/Ziegler/Byleen Finite Mathematics 11e Matrix Equations Let’s review one property of

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Basic Properties of Matrices Assuming that all

Barnett/Ziegler/Byleen Finite Mathematics 11e

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
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Barnett/Ziegler/Byleen Finite Mathematics 11e Basic Properties of Matrices (continued) Multiplication Properties

Barnett/Ziegler/Byleen Finite Mathematics 11e

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
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Barnett/Ziegler/Byleen Finite Mathematics 11e Basic Properties of Matrices (continued) Equality Addition:

Barnett/Ziegler/Byleen Finite Mathematics 11e

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

The use of these properties is best illustrated by an example of solving a matrix equation.
Example: Given an n x n matrix A and an n x p matrix B and a third matrix denoted by X, we will solve the matrix equation AX = B for X.

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Barnett/Ziegler/Byleen Finite Mathematics 11e Solving a Matrix Equation Reasons for each

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Example Example: Use matrix inverses to solve the system

Barnett/Ziegler/Byleen Finite Mathematics 11e

Example

Example: Use matrix inverses to solve the

system
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Barnett/Ziegler/Byleen Finite Mathematics 11e Example Example: Use matrix inverses to solve

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Example (continued) Form the matrix equation AX

Barnett/Ziegler/Byleen Finite Mathematics 11e

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:
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Barnett/Ziegler/Byleen Finite Mathematics 11e Example (continued) If the matrix A-1 exists,

Barnett/Ziegler/Byleen Finite Mathematics 11e

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:
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Barnett/Ziegler/Byleen Finite Mathematics 11e Example Solution The product of A-1 and

Barnett/Ziegler/Byleen Finite Mathematics 11e

Example Solution

The product of A-1 and B is

The solution

can be read off from the X matrix: x = 0, y = 2, z = -1/2
Written as an ordered triple of numbers, the solution is (0, 2, -1/2)
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Barnett/Ziegler/Byleen Finite Mathematics 11e Another Example Example: Solve the system on

Barnett/Ziegler/Byleen Finite Mathematics 11e

Another Example

Example: Solve the system on the right

using the inverse matrix method.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Another Example Example: Solve the system on

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Cases When Matrix Techniques Do Not Work

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.
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Barnett/Ziegler/Byleen Finite Mathematics 11e Application Production scheduling: Labor and material costs

Barnett/Ziegler/Byleen Finite Mathematics 11e

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?
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Barnett/Ziegler/Byleen Finite Mathematics 11e Solution Let x be the number of

Barnett/Ziegler/Byleen Finite Mathematics 11e

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.

This gives us the system of linear equations:
30x + 40y = 1800
20x + 30y = 1200
We can write this as a matrix equation: