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Matrix Revolutions: Solving Matrix Equations Matrix 3 MathScience Innovation Center Betsey Davis 00011000101001001110101001110001100010100100111010100111.

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Presentation on theme: "Matrix Revolutions: Solving Matrix Equations Matrix 3 MathScience Innovation Center Betsey Davis 00011000101001001110101001110001100010100100111010100111."— Presentation transcript:

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?


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