Math 3 Variable Manipulation Part 2 Systems with Matrices

1

Math 3 VM Part 2 Systems with Matrices September 12, 2018

Name:__________________________________ Date:________________________

Math 3 Variable Manipulation Part 2 Systems with Matrices

MATRICES An alternative method to solving system of equations is using Matrices. However, before we can solve systems of equations using matrices, you must have a basic understanding of matrices. A Matrix is an array of numbers: This matrix has 2 Rows and 3 Columns.

To add or subtract, just add or subtract the numbers in the same column and row and place answer accordingly.

𝟔 𝟐𝟒 𝟕

+ 𝟏 𝟑𝟓 −𝟐

= 𝟕 𝟓𝟗 𝟓

To multiply a matrix by a single number is easy:

We call the number ("2" in this case) a scalar, so this is called "scalar multiplication".

Multiplying a Matrix by Another Matrix But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns. To work out the answer for the 1st row and 1st column:

The "Dot Product" is where we multiply matching members, then sum up: (1, 2, 3) • (7, 9, 11) = 1×7 + 2×9 + 3×11 = 58 We match the 1st members (1 and 7), multiply them, likewise for the 2nd members (2 and 9) and the 3rd members (3 and 11), and finally sum them up. 1st row and 2nd column:

(1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139 And for the 2nd row and 2nd column: (4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 154

2


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

3


This may seem an odd and complicated way of multiplying, but it is necessary! Below is a real-life example to illustrate why we multiply matrices in this way. Example: The local shop sells 3 types of pies. Beef pies cost $3 each. Chicken pies cost $4 each. Vegetable pies cost $2 each. And this is how many they sold in 4 days:

Solution: Think about this ... the value of sales for Monday is calculated this way: Beef pie value + Chicken pie value + Vegetable pie value $3×13 + $4×8 + $2×6 = $83 So it is, in fact, the "dot product" of prices and how many were sold: ($3, $4, $2) • (13, 8, 6) = $3×13 + $4×8 + $2×6 = $83 We match the price to how many sold, multiply each, then sum the result. In other words: The sales for Monday were: Beef pies: $3×13=$39, Chicken pies: $4×8=$32, and Vegetable pies: $2×6=$12. Together that is $39 + $32 + $12 = $83 And for Tuesday: $3×9 + $4×7 + $2×4 = $63 And for Wednesday: $3×7 + $4×4 + $2×0 = $37 And for Thursday: $3×15 + $4×6 + $2×3 = $75 So it is important to match each price to each quantity. Now you know why we use the "dot product". And here is the full result in Matrix form:

They sold $83 worth of pies on Monday, $63 on Tuesday, etc. Rows and Columns To show how many rows and columns a matrix has we often write rows × columns. Example: This matrix is 2×3 (2 rows by 3 columns): When we do multiplication: The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix. And the result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix. Example: In that example we multiplied a 1×3 matrix by a 3×4 matrix (note the 3s are the same), and the result was a 1×4 matrix.

4


In General: To multiply an m×n matrix by an n×p matrix, the ns must be the same, and the result is an m×p matrix.

Order of Multiplication In arithmetic we are used to: 3 × 5 = 5 × 3 (The Commutative Law of Multiplication) But this is not generally true for matrices (matrix multiplication is not commutative): AB ≠ BA When we change the order of multiplication, the answer is (usually) different. Example: See how changing the order affects this multiplication

Sample Questions:

11. What is the matrix product [ 2𝑥3𝑥5𝑥

] [1 0 −1]?

a. [2𝑥 0 −2𝑥3𝑥 0 −3𝑥5𝑥 0 −5𝑥

]

b. [2𝑥 0 −2𝑥0 0 0

10𝑥 0 −10𝑥]

c. [2𝑥 3𝑥 5𝑥]

d. [9𝑥 0 −9𝑥]

e. [0]

12. A is a 3 × 2 matrix B is a 2 × 3 matrix C is a 2 × 2 matrix D is a 3 × 3 matrix Which of the following products does not exist?

a. AB b. AC c. BD d. CD

5


13.

14.

15.

16.

17.

18.

6


19.

20. Daisun owns 2 sportswear stores, (X and Y). She stocks 3 brands of T-shirts (A, B, and C) in each store. The matrices below show the number of each type of T-shirt in each store and the cost of each type of T-shirt. The value of Daisun’s T-shirt inventory is computer using the costs listed. What is the total value of the T-shirt inventory for Daisun’s 2 stores?

21. The number of people who shop at an electronics store during a given week is shown in the matrix below: Adolescents Adults Senior Citizens 75 100 30 The ratio of people from each age group who will purchase a product to the number of people in that age group who shop at the store is shown in the following matrix:

Adolescents 0.2 Adults 0.35 Senior Citizens 0.10

Based on the matrices, how many people will make purchases?

7


Identity Matrix The "Identity Matrix" is the matrix equivalent of the number "1":

A 3x3 Identity Matrix It is "square" (has same number of rows as columns). It has 1s on the diagonal and 0s everywhere else. Its symbol is the capital letter I. It is a special matrix, because when we multiply by it, the original is unchanged: A × I = A and I × A = A DETERMINANTS Determinants are like matrices, but done up in absolute-value bars instead of square brackets. There is a lot that you can do with (and learn from) determinants, but you'll need to wait for an advanced course to learn about them. However, below is the process to solve 2×2 and 3×3 determinants. (It is possible to compute larger determinants, but the process is much more complicated.) If you have a square matrix, its determinant is written by taking the same grid of numbers and putting them inside absolute-value bars instead of square brackets:

For the matrix: The determinant is: Just as absolute values can be evaluated and simplified to get a single number, so can determinants. The process for evaluating determinants is pretty messy, so let's start simple, with the 2×2 case. For a 2×2 matrix, its determinant is found by subtracting the products of its diagonals. In other words, to take the determinant of a 2×2 matrix, you multiply the top-left-to-bottom-right diagonal, and from this you subtract the product of bottom-left-to-top-right diagonal. Determinants are similar to absolute values, and use the same notation, but they are not identical, and one of the differences is that determinants can indeed be negative. Example: Evaluate the following determinant

Solution: multiply the diagonals, and subtract

Example: Find the determinant of the following matrix:

Solution: Convert from a matrix to a determinant, multiply along the diagonals, subtract, and simplify:

8


The computations for 3×3 determinants are messier than for 2×2's. Various methods can be used, but the simplest is probably the following:

Take a matrix A:

Write down its determinant:

Extend the determinant's grid by rewriting the first two columns of numbers:

Then multiply along the down-diagonals:

...and along the up-diagonals

Add the down-diagonals and subtract the up-diagonals:

And simplify:

Then det(A) = 1.

9


Example: Find the determinant of the following matrix:

Solution:

First convert from the matrix to its determinant, with the extra columns:

Then multiply down and up the diagonals:

Then add the down-diagonals, subtract the up-diagonals, and simplify for the final answer:

Note that your graphing calculator should be able to evaluate the determinant of any (square) matrix you enter. For instance:

But make sure, even if you have a graphing calculator, that you can evaluate 2×2 and 3×3 determinants, because you are likely to have word problems where the determinants contain variables that your calculator can't handle.

10


Sample Questions: Find each determinant

22.

23.

24.

25.

26.

27.

28.

29.

11


INVERSE MATRICE A matrices C will have an inverse C-1 if and only if the determinant of C is not equal to zero. (Since dividing by 0 is undefined.)

Where ad – bc is the determinant and is the adjoint of the matrix. Note: To find the adjoint, swap a and d and b and multiply b and c by -1.

Example: What is the inverse of A if A = [8 −5−3 2

]?

Solution: The inverse of a square matrix exists if and only if the matrix has a non-zero determinant. Find the determinant of A.

|A| = |8 −5−3 2

| = 16 – 15 = 1 ≠ 0 Therefore A-1 exists.

Example Questions: Find the inverse of each matrix

30.

31.

32.

33.

12


USING MATRICES WHEN SOLVING SYSTEM OF EQUATIONS Example: Consider the following simultaneous equations 3x+y=5 2x−y=0 Provided that we know how to multiply matrices we realize that our equations could be written as

First we find the inverse of the coefficient matrix:

The next step is to multiply both sides of our matrix equation by the inverse matrix:

Our solution is (1,2), the easiest way to check if we are right is to plug our values into our original equations. Example Questions: Use Matries to solve the system of equations

34. -2x _ 2y = 10 4x – 3y = -13

13


35. 5x + 2y = -15 -x + 4y = -19

36. -5x – 5y = 25 -2x - 4y = 16

37. Kara purchases 18 sets of hardware and 3 sheets of plywood for a skateboard ramp. Her friend purchases 5 sets of hardware and 5 sheets of plywood for a bike jump. If Kara spent $91.50 and her friend spent $65.00, how much do the sets of hardware cost? How much do the sheets of plywood cost?

38. Twice Jenny’s age plus five times Chloe’s age is 204. Nine times Jenny’s age minus three times Chloe’s is also 204. How old are Jenny and Chloe?

14


More Sample Questions:

39. The determinant of a matrix [𝑎 𝑏𝑐 𝑑

] equals ad – cb. What must be the value of x for the matrix [𝑥 8𝑥 𝑥

]

to have a determinant of -16?

40. Given that a[2 61 4

]

41. Find the inverse of the matrix:

[3 1−1 3

]

42. Determine A from the given expression

[−7 −66 5

]A = [4 2−5 0

]

43. Determine if X-1 exists.

X = [10 −5−4 2

]

15


Answers

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. A 12. CD

13. −2 98 8

14. 7

10

11 3

15. −7 31 −1−8 8

16. −3 82 −4

17.

18. .

19. .

20. . $7,350 21. 53

22.

23.

16


24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34. (-4, -1) 35. (-1, -5) 36. (-2, -3) 37. One sheet of plywood costs $9.50. 38. Jenny is 32 and Chloe is 28 39. x = 4 40. 8/3

41. 27[

3

10−

1

101

10

3

10

]

42. [

3

10−

1

101

10

3

10

]

43. None exists

Math 3 Variable Manipulation Part 2 Systems with Matrices

Documents

Transcript of Math 3 Variable Manipulation Part 2 Systems with Matrices