Download presentation

Presentation is loading. Please wait.

Published byElfreda Merritt Modified over 8 years ago

1
Matrix Revolutions: Solving Matrix Equations Matrix 3 MathScience Innovation Center Betsey Davis 00011000101001001110101001110001100010100100111010100111 00011000101001001110101001110001100010100100111010100111 00011000101001001110101001110001100010100100111010100111 00010010101001001110101001110001001010100100111010100111 00011010101001001110101001110001101010100100111010100111 1101100011101010101010011111011000111010101010100111 1101100011101010101010011111011000111010101010100111 1101100011101010101010011111011000111010101010100111 1101101011101010101010011111011010111010101010100111 1101100011101010101010011111011000111010101010100111 111101010101010101010101010100111111101010101010101010101010100111 111101010101010101010101010100111111101010101010101010101010100111

2
Matrix Revolutions B. Davis MathScience Innovation Center It’s time to use matrices! How would you solve 3 x = 6 for x? We need to think of multiplying, not dividing because with matrices there is no “divide”. Multiply 3x and 6 by the multiplicative inverse of 3. (1/3)3x = (1/3)6 so…. X = 2

3
Matrix Revolutions B. Davis MathScience Innovation Center The Big Idea If [A] [x] = [B] where A, B, x are matrices, Then [A] -1 [A] [x] = [A] -1 [B] So [x] = [A] -1 [B]

4
Matrix Revolutions B. Davis MathScience Innovation Center (1/3)3x = (1/3)6 Multiplication by real numbers is commutative, so order is not important. Multiplication by matrices is NOT commutative, so order is VERY important. Let’s solve for x: X =

5
Matrix Revolutions B. Davis MathScience Innovation Center First, find the inverse of the left matrix X =

6
Matrix Revolutions B. Davis MathScience Innovation Center Second, multiply both sides of the equation by A -1 X =

7
Matrix Revolutions B. Davis MathScience Innovation Center Third, simplify both sides. X =

8
Matrix Revolutions B. Davis MathScience Innovation Center To simplify the right hand side, multiply the 2 matrices. X =

9
Matrix Revolutions B. Davis MathScience Innovation Center By Calculator: X = So, just enter A and B into the calculator. Then on the home screen type [A] x -1 [B] enter.

10
Matrix Revolutions B. Davis MathScience Innovation Center X = Final Answer for x.

11
Matrix Revolutions B. Davis MathScience Innovation Center SO WHAT??????? We can use this new skill– solving equations using matrices – To solve linear systems.

12
Matrix Revolutions B. Davis MathScience Innovation Center Let’s learn how! Basic idea comes from solving AX = B Let’s write a system: 3 x + 2 y = 6 5 x - 9 y = 15 Now, let’s rewrite the system using matrices:

13
Matrix Revolutions B. Davis MathScience Innovation Center 3 x + 2 y = 6 5 x - 9 y = 15 Re-writing the system using matrices: Make a matrix of coefficients. Make a matrix of variables. Make a matrix of constants. =

14
Matrix Revolutions B. Davis MathScience Innovation Center 3 x + 2 y = 6 5 x - 9 y = 15 What size are these and can they be multiplied? What size is the answer? = 2 x 22 x 1

15
Matrix Revolutions B. Davis MathScience Innovation Center 3 x + 2 y = 6 5 x - 9 y = 15 We know we can multiply them and the answer is a 2x1. We will use the same BIG IDEA: If [A][x]=[B], then [x] = [A] –1 [B] =

16
Matrix Revolutions B. Davis MathScience Innovation Center Here we go! = Remember BIG IDEA: If [A][x]=[B], then [x] = [A] –1 [B] Important!!! Notice order of multiplication

17
Matrix Revolutions B. Davis MathScience Innovation Center We have identity matrix on left =

18
Matrix Revolutions B. Davis MathScience Innovation Center The identity times [x] is [x]. = Now just type [A]-1[B] on your TI

19
Matrix Revolutions B. Davis MathScience Innovation Center = What does this mean? The solution is the ordered pair Final answer

20
Matrix Revolutions B. Davis MathScience Innovation Center Let’s try it ! 2w – x + 5 y + z = -3 3w + 2x + 2 y – 6 z = -32 w + 3x + 3 y - z = -47 5w – 2 x - 3 y + 3 z = 49 We will need 3 matrices…

21
Matrix Revolutions B. Davis MathScience Innovation Center Matrix of Coefficients 2w – 1 x + 5 y + 1z = -3 3w + 2x + 2 y – 6 z = -32 1w + 3x + 3 y -1 z = -47 5w – 2 x - 3 y + 3 z = 49

22
Matrix Revolutions B. Davis MathScience Innovation Center Matrix of variables 2w – x + 5 y + z = -3 3w + 2x + 2 y – 6 z = -32 w + 3x + 3 y - z = -47 5w – 2 x - 3 y + 3 z = 49

23
Matrix Revolutions B. Davis MathScience Innovation Center Matrix of constants 2w – x + 5 y + z = -3 3w + 2x + 2 y – 6 z = -32 w + 3x + 3 y - z = -47 5w – 2 x - 3 y + 3 z = 49

24
Matrix Revolutions B. Davis MathScience Innovation Center =

25
=

26
=

27
=

28
=

29
Matrix Revolutions B. Davis MathScience Innovation Center = Therefore, The solution is An ordered quadruplet: (2,-12,-4, 1)

30
Matrix Revolutions B. Davis MathScience Innovation Center Way cool, huh?

Similar presentations

© 2023 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google