Math-Precalculus

Matrix methods can be used to solve linear programming problems . A linear programming problem is used to find an optimal solution , subject to stated restraints

One typical application is to maximize profits . For example , a beauty parlor provides both highlighting and permanent wave services . It costs 5 in materials and requires 30 minutes to provide highlighting However , it costs 12 in materials but requires 80 minutes to provide a perm . The store has at most 120 in materials and 800 minutes in labor per day to expend

How many highlighting services and

how many perms can the beauty parlor perform daily

x Highlights , y perms Multiply row 1 by -6 and add it to row 2 Multiply row two by 1 /8 Multiply row 2 by -12 and add it to row 1 Divide row 1 by 5 x 0 ,y 10

So , the answer for optimum use of time and product is : do no highlights and do 10 perms Red line is the equation x 0 (which is the y axis , Black line is y 10

You are given the following system of linear equations

3x - 2y z 2

-x y 3

-2y 6 -1

Provide a coefficient matrix corresponding to the system of linear equations What is the inverse of this matrix

Multiply Row 1 by 1 /3 . Then add it to row 2

Multiply row 2 by 2 and add it to row 1

Multiply Row 2 by 6 and add it to row 3

Divide Row 3 by 2

Multiply Row 3 by -1 and add it to row 1

Multiply Line 2 by 3

Subtract line 3 from line 2 What is the transpose of this matrix Find the determinant for this matrix Calculate the following for C A B -3A A-1

Divide row 3 by -2 and add it to row 2

Multiply Row 2 by 2 /3

Multiply Row 2 by -2 and add it to row 1

Multiply row 1 by -8 , add it to row 3

Multiply row 2 by -9 , add it to row 3

Divide row 3 by -10

Multiply row 3 by -1 , add it to row 2 and Row 1 Solve the following linear system using Gaussian elimination

Show work

2y z 4

x y z 6

2x y z 7

Swap Row 2 and Row 1

Divide row 2 . Then multiply row 2 by -1 and add it to row 1

Divide row 3 by 2 . Then multiply row 1 by -1 and add it to row 3

Divide row 2 by -2 and add it to row 3

Multiply Row 3 by -4

Therefore , z 6 The solution is x 1 , y -1 , z 6

Solve the following linear system for x using Cramer 's rule

Show work

4x - y z -5

2x 2y 3z 10

5x - 2y 6z 1 Thus , the solution is x -1 , y 3 , z 2...