unit 3: matrices - ellitech.caellitech.ca/mathinservice/b30mathtext/unit 3 matrices.doc · web...

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Unit 3: Matrices Objectives: C.1 To define basic terms associated with matrices C.2 To create a matrix to illustrate a given situation. C.3 To add and Subtract Matrices C.4 To add and subtract matrices using scalar multiplication. C.5 To multiply two matrices. (not larger than 3×3) C.6 To determine the properties of matrices with respect to addition, scalar multiplication and matrix multiplication. C.7 To use row operations with matrices C.8 To determine the inverse of a 2×2 matrix. C.9 To solve matrix equations using multiplication by an inverse. Matrix Terminology A matrix is a rectangular array of numbers. For Example: . , , , and All of the examples above are matrices (plural). Most texts will enclose matrices in square brackets. Each number in a matrix is called an Entry or Element of that matrix. They are arranged in Rows (horizontal) and Columns (vertical). The number of rows and columns, in that order, make up the dimensions of the matrix (r c). Matrices are usually named using capital letters. In the example above; A is a (2 3) matrix (read 2 by 3), R is a (1 3) matrix, C is a (3 1) matrix, S is a (3 3) matrix, and O is a (2 2) matrix. This information is often summarized in subscripts; A 23 , R 13 , C 31 , S 33 , and O 22 Lower case letters with subscripts may be used to refer to individual entries in a matrix. For matrix N the element in row r and column c is referred to as n rc . -45-

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Page 1: Unit 3: Matrices - Ellitech.caellitech.ca/MathInservice/B30MathText/Unit 3 Matrices.doc · Web viewIt may be necessary to simplify the elements within a matrix to determine if corresponding

Unit 3: MatricesObjectives:C.1 To define basic terms associated with matricesC.2 To create a matrix to illustrate a given situation.C.3 To add and Subtract MatricesC.4 To add and subtract matrices using scalar multiplication.C.5 To multiply two matrices. (not larger than 3×3)C.6 To determine the properties of matrices with respect to addition, scalar multiplication and

matrix multiplication.C.7 To use row operations with matricesC.8 To determine the inverse of a 2×2 matrix.C.9 To solve matrix equations using multiplication by an inverse.

Matrix TerminologyA matrix is a rectangular array of numbers.

For Example:

. , , , and

All of the examples above are matrices (plural). Most texts will enclose matrices in square brackets.

Each number in a matrix is called an Entry or Element of that matrix. They are arranged in Rows (horizontal) and Columns (vertical). The number of rows and columns, in that order, make up the dimensions of the matrix (r c). Matrices are usually named using capital letters.

In the example above;A is a (2 3) matrix (read 2 by 3), R is a (1 3) matrix, C is a (3 1) matrix, S is a

(3  3) matrix, and O is a (2 2) matrix.

This information is often summarized in subscripts; A23, R13, C31, S33, and O22

Lower case letters with subscripts may be used to refer to individual entries in a matrix. For matrix N the element in row r and column c is referred to as nrc.

For Example: Matrix Nab:

In Matrix A, above:

a11 = 9, a12 = -3, a22 = 1, etc

a32 = Ø because there is no third row.

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Unit 3: MatricesA matrix with only 1 row is called a Row Matrix. is a row matrix.

A matrix with only 1 column is called a Column Matrix. is a column matrix.

A matrix with the same number of rows and columns is called a Square Matrix.

is a square matrix of order 3, and

is a square matrix of order 2.

A matrix whose entries are all zeros is called a Zero Matrix. O2 is a zero matrix. ‘O ab’ is usually used to denote a zero matrix.

Two matrices are equal iff (if and only if) they have the same dimensions and the same corresponding entries. For example;

but and

It may be necessary to simplify the elements within a matrix to determine if corresponding entries are equal:

For every matrix A there is a Transpose matrix written AT which is formed by interchanging the rows and columns of matrix A.

If , then

Notice that the dimensions and the designation of individual entries will be the reverse for a transpose matrix.

For the matrix A23, the corresponding transpose matrix is and for the entries of the matrices;

and for

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Unit 3: MatricesPractice Questions 1: Matrix Terminology

1. For the Following matrices;

, , , and

a. State the dimensions of each matrix using subscripts (Nrc)b. How many elements are in each matrix.c. List the elements in the third column.d. List the entries in the first row.e. List the following elements, if possible; a22, c21, d32, and b24.

2. Choose a capital letter for each of the following and name it using subscripts for the dimensions.

a. b.

c. d.

e. f.

3. Find the values of x, y, and z that will make each pair of matrices equal.

a. b.

c. d.

4. For every matrix in question 1 a.write the transpose matrix. b. name the corresponding entries from 1. e.

5. State the dimensions of the transpose of the matrices in question 2.

6. Use a matrix to illustrate the following class enrollment;

Class No. of Students Average Std. DeviationHistory 27 69.5 8.7English 31 71.2 9.5

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Unit 3: MatricesAddition, Subtraction, and Scalar Multiplication of Matrices

Matrices of the same dimension can be added by adding together the corresponding entries. Only matrices with the same dimensions can be added.

For Example:

For every matrix A there is an Additive Inverse(negative), -A, for which A + (-A) = O. Each entry of -A is the Additive Inverse (negative) of the corresponding entry of A.

For Example:

If , then , and A + -A =

Subtraction of matrices is equivalent to adding the negative of a matrix. A - B = A + (-B)

For Example:

Scalar Multiplication is the multiplication of a scalar (real number) and a matrix. The product can be found by multiplying every element of the matrix by the scalar. Scalar multiplication is commutative, (r A) = (A r), and associative,r (s A) = (r s) A.

For Example:

Division of a matrix by a scalar is best approached as multiplication by the inverse of the scalar.

For Example:

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Unit 3: MatricesPractice Questions 2: Addition, Subtraction and Scalar Multiplication of Matrices

1. Add or subtract the following(if possible);

a. b.

c. d.

e.

2. Find the product of each of the following;

a. b.

c. d.

3. Given the matrices;

, , and

Calculate the following;

a. A - 2B b. 2(A + B)c. 2A + 2B d. C - AT

e. 2B + CT

4. Given the following matrices;

, , and

Prove:

a. L + M = M + L b. (q ×r)N = q(r ×N)c. t × (M - N) = tM - tN

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Unit 3: MatricesBuilding Matrices

In many fields of work and study, it is convenient to use matrices to organize data.

For example, Krazy Wally’s Appliances and Pizza Boutique uses Matrices to keep track of the sales made by each salesperson.

Wally's Sales Anita's SalesItem January February March Item January February March

Andywood J200 Tape Deck 7 3 9 Andywood J200 Tape Deck 12 7 9Panatronic 24 in. TV 2 7 11 Panatronic 24 in. TV 5 11 7Steer's Best Microwave 5 3 5 Steer's Best Microwave 3 4 12Extra Large Loaded Pizza 1 1 6 Extra Large Loaded Pizza 8 7 14Catsup & Mustard Pizza 16 3 7 Catsup & Mustard Pizza 0 1 2

a. Make a matrix to show the units moved by each salesperson.b. Find the total sales for each month by adding the matrices.c. How many more sales than Wally did Anita Make in each category?

Solution:a. Using the rows to represent the items and the columns to represent the months;

Wally’s Sales are, and Anita’s sales are,

b. The Total sales can be found by calculating W + A;

c. We can find out how many more sales Anita made by calculating A - W;

In the example above we can indicate each entry using subscript notation;Anita’s sales of Catsup and Mustard Pizza in February are; a52 = 1. Wally’s sales of the same item for the same period are; w52 = 3

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Unit 3: MatricesThe possible routs through a Network can be illustrated using a matrix;

To:

From:

In the matrix at the right, a 0 indicates no path from the first point to the second, a number indicates the number of routs. Row 1 shows routs from point A. Column 1 Shows Routs to Point A. The 2 in row 3 column 4 indicates that there are two routs from point C to point D.

Although the table is easier to read, the matrix is a form that is more compact and easier to enter into calculations.

Network Matrices can be used to describe electrical networks, telephone networks , and computer networks.

An Incidence Matrix can be used to summarize the line segments and points of a geometric figure;

In an incidence matrix, rows represent line segments and columns represent points. If a point is contained in a particular line segment, a 1 is entered in that segment’s row at the point’s column. If the point is not contained in the segment, a 0 is entered.

For Example: Parallelogram ABCD can be represented as follows;

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From A to A = 0 From A to B = 1 From A to C = 0 From A to D = 0From B to A = 1 From B to B = 0 From B to C = 1 From B to D = 0From C to A = 0 From C to B = 1 From C to C = 1 From C to D = 2From D to A = 1 From D to B = 0 From D to C = 1 From D to D = 0

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Unit 3: MatricesPractice Questions 3: Building Matrices

1. The following tables show the consumption of energy, by type, in Canada from 1991 to 1994. Use this information to answer the following;

CoalTotal Use × 1015 Joules

Year 1991 1992 1993 1994Producers' own consumption 5.0 2.7 3.9 3.9

Non-energy use 10.0 6.2 6.5 9.4Energy use final

demand 185.3 187.1 179.8 166

Industrial 183.1 185.1 177.9 164.3Transportation 0.0 0.0 0 0Agricultural 0.0 0.0 0 0Residential 2.1 2.0 1.8 1.5

Public administration 0.1 0.0 0 0.2Commercial and

institutional 0.0 0.0 0 0

ElectricityTotal Use × 1015 Joules

1991 1992 1993 1994Producers' own consumption

136.2 134.4 132.3 143.9

Non-energy use 0 0 0 0Energy use – final demand

1570.7 1593.6 1625.1 1640.2

Industrial 667.4 673.4 693.1 712.8Transportation 11 12.3 13.1 13.3Agricultural 34.3 34.1 34.1 34.8Residential 461.2 471.9 477.7 470.7Public administration 47 47.3 47 47.1Commercial and institutional

350 354.6 360 361.6

a. Express the information in each table using a matrix. (Do not include the year.)b. Use matrix addition to find and express the total energy used from coal and electricity.c. Use matrix addition to calculate how much more electricity was used in each category.

2. Draw a network for the following matrix;

3. Write a matrix to describe the following network;

4. Make an incident matrix for the following figure;

Q

R

P

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Unit 3: Matrices5. Use the information in the following table to create two matrices, for Canada and for

Sask.Use matrix arithmetic to create a matrix showing the population outside Saskatchewan.Population By Aboriginal Groups (1996 Census)

Canada

Aboriginal groups TotalAge groups

Under 15years

15-24years

25-44years

45+years

Total population 28 528 125 5 899 200 3 849 025 9 324 340 9 455 560Non-Aboriginal population 27 729 115 5 618 785 3 705 230 9 080 735 9 324 365Aboriginal population 799 005 280 420 143 795 243 610 131 185

Saskatchewan

Aboriginal groups Total -Age groups

Under15 years

15-24years

25-44years

45years +

Total population 976 615 228 480 141 190 284 715 322 225Non-Aboriginal population 865 370 182 125 120 685 255 190 307 370Aboriginal Population 111 245 46 360 20 510 29 525 14 855

Matrix Multiplication

The school canteen sells sandwiches for $2.50, potato chips for $0.75, and chocolate bars for $1.00. This information can be expressed in matrix In January, they sold 120 sandwiches, 230 bags of potato chips and 175 chocolate bars. In February, they sold 95 sandwiches, 195 bags of chips, and 150 chocolate bars.

This could be expressed in a matrix:

To find the gross sales of the canteen in each month, we would multiply the price for each item by the number of that item sold and add their products:

This exercise provides us with a reasonable method for multiplying matrices. Notice that we multiplied the elements of each row of C by the elements of each column of A and added their products. Also notice that the result was a 1 × 2 matrix.

In general, matrices can only be multiplied if the number of columns of the first matrix is the same as the number of rows in the second. The product is a matrix with the the same number of rows as the first and the same number of columns as the second.

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Unit 3: MatricesRemember, in order to multiply two matrices their dimensions must match this pattern:

For Example: The multiplication of two 2 × 2 matrices;abcd

wxyz

aw by

abcd

wxyz

aw

bby ax bz

abcd

wxyz

aw by ax bzcw dy

a

bbcd

wxyz

aw by ax bzcw dy cx dz

The result is another 2 × 2 matrix.

For Example:Given the following matrices;

, , and

Calculate the following, if possible;a. AB b. BAc. CA

Solution: a. The product AB can not be found because A has 3 columns and B has 2 rows.

b. B× A =

3691 2

12 3456

31 64 27

3691 2

12 3456

27 32 65 27 36

3691 2

12 3456 6

27 36 33 66 27 36 45

3691 2

12 3456

27 36 4591 12 4

27 36 4557

3691 2

12 345 56

27 36 4557 921 25

27 36 4557 78

3691 2

12 3456

27 36 4557 78 931 26

27 36 4557 78 99

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Unit 3: Matricesc.

Notice that matrix multiplication does not obey the Commutative Property of Multiplication. Unlike multiplication of real numbers, for matrices:

A ×B ≠ B×A

Matrix multiplication does follow the Associative and Distributive Properties;

(A ×B) × C = A × (B×C)

and

A × (B + C) = AB + AC

Practice Questions 4: Multiplying Matrices

1. Given the following matrices use their dimensions to state which of the following multiplications are possible. If possible, list the dimensions of the product (P).

a. AB b. BAc. AD d. DAe. AE f. EAg. BE h. EBi. BD j. ECk. A2 l. C2

2. Find the product, if possible, for each of the following:

a. b.

c. d.

e. f.

3. Given the following matrices;

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Unit 3: Matrices

Calculate the following;a. QS b.SQc. RS d.SRe. QR f. RQg. What conclusions can we draw about matrix S?h. What about Q & R?

4. A canoe manufacturer builds Cedar Strip Canoes, Wood and Canvas canoes and Fiberglass canoes. Her cost per canoe is $275 for the cedar strip, $540 for the wood and canvas, and $210 for the fiberglass. The following chart shows the number of canoes she has made for the last two years:

Canoe 1996 1997Cedar Strip 6 8

Wood & Canvas 3 5Fiberglass 12 9

a. Express the cost and the data from the table in two matrices.b. Use matrix multiplication to find her total costs per year.

TI-82 Calculator

The TI-82 calculator is able to perform matrix calculations using the [MATRX] key. To enter the matrix:

-Press [MATRX] [→][→] (arrow to EDIT menu) Choose 1 of [A], [B], [C], [D], or [E]-Enter the dimensions of the matrix, pressing [ENTER] after the rows and columns.-Enter the elements of the matrix, pressing [ENTER] after each one.-Repeat the process to enter more matrices for up to 5 in total.or press [2nd]-[MODE] to quit.

To use matrices in calculation;-Press [MATRX] and choose one of the names given to insert that matrix into your

calculations. The result will be shown on the screen enclosed in square brackets.-To store the results of a calculation press [STO→] and then use the [MATRX] key to

choose a name.

Note: If you attempt an operation that is not possible because the matrices have the wrong dimensions, the calculator will show the message: “ERR: DIM MISMATCH”

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Unit 3: MatricesThe Determinant ()

For every square matrix, there is a real number called the determinant calculated by subtracting the sum of the products of the right to left diagonals from the sum of the products of the left to right diagonals.

For example in the 2× 2 matrix

Aabcd

In some texts, the determinant of a as written;

In Larger matrices the following pattern may be useful:

TI-82 Calculator The determinant for any matrix can be calculated by pressing [MATRX] [→] (math menu) , Choose 1:det. Then enter the matrix name from the matrix menu.

Example: Find the determinant for ;

Solution:

Example: Evaluate;

Solution:

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Unit 3: MatricesPractice Questions 5: The Determinant

1. Find the determinant of each of the following:

a. b.

c. d.

e. f.

g. h.

2. Given the following matrices;

, and

And the number, n. Prove the following:

a. (nA) = n2 (A) b. (BT) = (B)

c. (AB) = (A)•B) d. (A - AT) = ( AT - A)

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Unit 3: MatricesInverse of Matrices

In the multiplication of real numbers, the number 1 is called the Identity because multiplication by 1 does not change the value of the number being multiplied:

3×1 = 3, 12 ×1 = 12, x × 1 = x

For square matrices, the Identity Matrix (Inn) has ones in the diagonal from upper left to lower right, and zeros as all other entries:

, , and

These matrices have the property that for all square matrices .

For Example: The multiplication:5791 1

1001

51 70

5791 1

1001

55 07 1

5791 1

1001

5791 11 0

5791 11

1001

5799 01 11

5791 1

Shows that the matrix is the identity for a 2 × 2 matrix.

The diagram on the next page shows that the matrix is the identity for a

3 × 3 matrix.

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Unit 3: Matrices

For Example: The 3× 3 identity matrix in action:

12 345678 9

10 001 000 1

11 20 30

1

12 345678 9

10 001 000 1

11

02 13 01 2

12 345678 9

10 001 1000 1

12 10 20 31 12 3

12 34

55678 9

10 001 000 1

12 341 50 60

12 34

12 345678 9

10 001 000 1

12 344 05 16 0

12 345

12 345678 9

10 001 000 1

12 345 40 50 61

12 3456 6

12 345678 9

10 001 000 1

12 345 6

771 80 90

12 34567

12 345678 9

10 001 000 1

12 345677 08 19 0

12 3456778

12 345678 9

10 001 000 1

12 345

66

787 08 09 1

12 345678 9

Once again we find that the product is identical to the original matrix.

In the future, unless the full multiplication is requested, we can assume that the product of a matrix and the identity is equal to the original matrix.

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Unit 3: Matrices

The multiplicative inverse of a real number is the number which when multiplied by the original number produces the product 1.

; is the inverse of 2 (and visa versa).; is the inverse of x (and visa versa).

The multiplicative inverse of x is generally written as or x-1.

The inverse of a square matrix is the matrix that, when multiplied by the original matrix, produces the identity matrix (I).

If , then A and B are inverses. B = A-1 and A = B-1.

In general, we can find the inverse of any 2 2 matrix using the following formula;

If , then

For Example: Find the inverse of .

Solution: Find the determinant,

Fill in the formula for the inverse. The inverse of the determinant times the matrix; interchange upper right with lower left and make the other entries negative.

Check by multiplying:

NB. If (A) = 0, there is no inverse for the matrix.

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Unit 3: MatricesPractice Questions 6: Inverse of a 2× 2 Matrix.1. Find the inverse of each matrix and check by multiplying. If the inverse can not be found

explain.

a. b.

c. d.

e. f.

g. h.

2. a. Use your answer for 1 a. to prove that . (Multiplication by the inverse is commutative.)

b. Use the answers for 1. a. and d. to prove that

Using Inverses to Solve Matrix Equations

In earlier math courses we have used multiplicative inverses to solve equations:3x = 21

x = 7We can use the inverse of a matrix in a similar way to solve matrix equations:

For Example: Solve for A.

Solution: First, find the inverse of :

Next, left multiply both sides of the equation by the inverse:

Check by multiplying:

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Unit 3: MatricesPractice Questions 7: Solving Matrix Equations

1. Solve each of the following for A.

a. b.

c. d.

For e. and f. use addition or subtraction to isolate the term with A in it before solving.

e. f.

Solving Systems of Equations Using Inverse Matrices

In earlier courses we learned how to solve systems of linear equations by three methods;

1. Graphing- Draw the graph of the equations and find the intersection.

For Example: Find the solution for -2x + y = 7, and x - 5y = 10

Solution: Graph both equations on the same set of axes. Find the intersection point.Solve both equations for y and make a

table of values:

y = 2x + 7x -7 -3 -2y -7 1 3

x -5 0 5y -3 -2 -1

Check by substituting into the original equations:

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Unit 3: Matrices

-2(-5) + (-3) ?

7 (-5) - 5(-3) ?

 10

10 - 3  ?

7 -5 + 15 ?

 107 = 7 _ 10 = 10 3

The solution set is (-5, -3)

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Unit 3: Matrices2. Substitution: -Solve one equation for x or y and substitute into the second.

For Example: Find the solution for -2x + y = 7, and x - 5y = 10

Solution: Solve the first equation for y: y = 2x + 7Substitute into the second equation and solve: x - 5(2x + 7) = 10 x - 10x - 35 = 10 -9x = 45 x = -5Substitute back into the first equation: -2(-5) + y = 7 10 + y = 7 y = -3

The solution set is (-5, -3). The check is the same as for the previous example.3. Linear Addition or Subtraction: Add and subtract multiples of the equations together to

eliminate one of the variables.

For Example: Find the solution for -2x + y = 7, and x - 5y = 10

Solution: -2x + y = 72( x - 5y = 10 )

-2x + y = 7+(2 x - 10y = 20 )

-9y = 27y = -3

-2x + (-3) = 7-2x = 10

x = -5

- Write the equations above each other.- Multiply one of the equations by a constant to make one of the coefficients match

- Add (or subtract) to eliminate the matching variable.

- Solve for the variable.- Substitute and solve for the remaining variable.

The solution set is (-5, -3). Again, the check is the same as for the first example.

We can use the inverse matrix to solve systems of equations. This takes advantage of the property that, if A•X = B then X = A-1•B

For Example: Find the solution for -2x + y = 7, and x - 5y = 10

Solution: First, write the system using matrices:

Next, find the inverse; (see page 56)

Finally, solve using matrix multiplication:

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Unit 3: Matrices

The solution set is (-5, 3). Once again, the check is as in the first example. TI-82 Calculator: To find the inverse of a matrix;

- Enter the matrix into matrix [A] using the [MATRX] EDIT menu.- Exit to the main screen by pressing [2nd]-[MODE]- Enter [A]-1 using the [MATRX] and [x-1] keys and press [ENTER]

You can see the elements in fraction form by pressing [MATH], choose 1:Frac before pressing [ENTER]

NB If the matrix has no inverse, then the matrices are either inconsistent (no solution) or dependent (infinite solutions)

For Example: Solve the system of equations: x + 2y = 8-2x + 4y = 10

Solution: First, write the equations in matrix form;

Next, find the inverse;

Use the inverse to solve the equation;

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Unit 3: Matrices

Check by substitution: 3

The solution set is .

TI-82 Calculator: To solve the equations above, enter into [A] and into [B] then

key in the following: [A]-1*[B] and press [ENTER].

If you wish to see the solution matrix in fraction form use [MATH] 1:Frac

Practice Questions 8: Solving Systems of Equations Using Inverse Matrices

1. Solve the following systems of equations, if possible, using inverse matrices;

a. -2x + 5y = 4 b. 5x - 3y = 11x - 2y = -1 3x + y = -13

c. 6x + 4y = 4 d. 7x + 5y = -5-9x + 6y = 0 -2x + 2y = 22

e. 12x - 15y = -1 f. 3x + 2y = 52x + 6y = 3 -6x - 4y = -10

g. 5x - y = 3 h. 2x - y = -8-5 + y = -2 -3x + 2y = 21

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Unit 3: Matrices2. A farmer has two grain trucks but he does not know how much grain each one can haul. One

day he took 6 loads with one truck and 3 loads with the other and was paid for 39 tonnes of wheat. The next day he took 4 loads with the first truck and 5 loads with the second and was paid for 41 tonnes of wheat. Write equations using two variables to describe the situation and use matrices to find the capacity of both trucks.

Using Row Operations to Solve Systems of Equations

In the previous section, we reviewed the process of linear addition and subtraction used to solve systems of equations. We can use similar skills with matrices to solve systems of linear equations:

For Example: Find the solution for -2x + y = 7, and x - 5y = 10 (yet again)

Solution: Augmented Matrix

2nd row multiplied by 2

1st row added to 2nd

1st row minus 2nd.

1st row multiplied by

The matrix on the bottom at right corresponds to or

We have solved the system of equations using Row Operations.

There are three Elementary Row Operations:1. Interchange any row with another row.2. Multiply any row by a constant.3. Replace any row by adding (or subtracting)

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Unit 3: Matrices any other row.

For Example: Use an augmented matrix to solve the following system of equations:4x + y + z = 53x + 3y - 2z = 22

x - 2y - z = 3

Solution: Set up an augmented matrix using the coefficients from the equations.

Row 1 subtract row 2

Row 1 subtract row 3

Row 1 multiplied by

Row 2 subtract row 1

Row 2 multiplied by 1/3

Row 2 minus row 3

Row 2 multiplied by 1/3

Row 3 plus row one

Row 3 plus 2 times row 2

Reading the final matrix, we can see that:Z=-5, Y=2, X=2Purists may want to switch rows one and three to have the solutions in alphabetical order.

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Unit 3: MatricesFor Example: Solve the following system of equations;

3x + 5y - z = 3-2x -3y +3z = -6

-x + 4z = -5

Solution:Set up an augmented matrix using the coefficients from the equations.Notice that in the third row a zero is tje coefficient of y.

Row 1 plus row 2

Row 1 plus row 3

Row 1 multiplied by

Row 2 subtract two times row 3

Row 2 plus three times row 1

Row 2 times

Row 1 minus three times row 2

Row 3 minus four times row 2

Row 3 times -1

- When placed in order, the matrix shows that: x = -3, y = 2, and z = -2

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Unit 3: MatricesTI-82 Calculator can solve this system using the inverse.

Store the matrix as [A], and the matrix as [B].

Then enter the following calculation using the [MATRX] and [x-1] keys;

[A]-1*[B] and press [ENTER].

The calculator will display the solution as the 3 × 1 matrix;

Practice Questions 9: Solving Systems of Equations Using Row Operations

1. Use row operations on the matrix to produce the following results. Start

with the original matrix in each case.

a. First row first element to be a 1 b. First row second element to be a 1c. Second row first element to be a 1 d. Produce two zeros in one row.e. Produce two zeros in two rows.

2. Use augmented matrices to solve the following systems of equations:

a. x + 2y = -1 b. 2x + 3y = 152x - y = 13 -3x - y = 2

c.x + y + z = 6 d. x - y + z = 32x - y - z = -3 3x + 2y -z = 1

x - 3y + 2z = 1 4x - 2y - 3z = -2

e. 3x + 5y = 6 f 3x - 2y = 4-4x + 2y = 5 -6x + 4y = -3

g. 1y - 4z = -10 h. 5x + 3y = -22x + y = -1 4 x + 7z = 103x + 2z = -2 8y - 3z = 2

i. 9x - 5y + 2z = 9 j. 4x + 5y + z = -23x + 2y + 3z = 0 6x - 2z = -11

-2x + y - 4z = 19 4y - 3z = -3/5

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Unit 3: MatricesInequalities and Linear Programming

Objectives:

C.10 To graph systems of inequalities.C.11 To determine the intersections of the lines graphed for systems of inequalities.C.12 To determine which vertices of the polygon formed by systems of inequalities will

maximize or minimize a given linear function. (Linear Programming)

Notes:Graphing Inequalities

In Mathematics 10, we solved and graphed simple inequalities with one variable;

For example:

Remember that multiplying or dividing by a negative number reverses the sign.

The solution is often displayed on a graph.

Inequalities in two variables can be displayed on the Cartesian Coordinate Plane;

For Example: 2x - y < 3 2x < y + 3 Solve for y2x - 3 < y

y > 2x - 3 Expressed with y first.

Create a table of values and graph.

x -2 -1 0 1 2y -9 -5 -3 -1 1

Use a test point to find which side of the line is to be shaded.

Test Point (1, 1) y > 2x - 31 > 2(1) – 31 > -1 3

If the sign is < or > use a dashed line.

If the sign is ≤ or ≥ use a solid line.

Graph of y > 2x - 3

Generally, when the equation is solved for y,

- a > or _ sign will result in the area above the line being shaded, - a < or _ sign will result in the area below the line being shaded.

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Unit 3: MatricesTI-82 Calculator: The calculator will not graph inequalities but it will allow the use of the

DRAW menu. For best results, clear the [y=] menu and press [ZOOM] 6:ZStandard before using the DRAW functions.

The following keystrokes will display the graph of y > 2x -3:

- [2nd]-[PRGM] (draw menu) choose 7:Shade( - Enter the following; Shade(2x-3,10,2) and press [ENTER] the graph will be displayed.

The argument within the shade command goes; Shade(lower function , upper function , resolution). In our example, we shaded the area

between y = 2x - 3 and y = 10 with lines 2 pixels apart.

To display the graph of y < 2x - 3 use the DRAW menu ([2nd]-[PRGM]) to enter the following; Shade(-10, 2x-3, 3) and press [ENTER]

To clear the drawing press [ZOOM] 6 (standard)

Practice Questions 10: Graphing Inequalities

1. Graph each of the following equations on the Cartesian plane:

a. 3x - 2 = y b. y = 7c. x = -3 d. 2x - 3y = 6e. 3x + 4y = 9 f. 5y + 3x = 7

2. Graph the solution for the following inequalities:

a. y > 2(y - 3) b. x + 2 ≤ 2(x - 3)c. 2x + y ≥ 4 d. y > x - 3e. 3y - x ≤ x + 2 f. 6x - 4y < 2(6 - y)

3. For each of the following diagrams write the inequality that describes the shaded region:

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Unit 3: MatricesSystems of inequalities

Once we have mastered the skill of graphing linear inequalities, we can expand it to include determining the area designated by the intersection of two or more inequalities together;

For Example: Graph the area defined by the intersection of;x + 2y < 63x - y > 6y ≥ x - 8

Solution: Solve each equation for y and set up a table of values for the lines.

x + 2y < 62y < -x + 6y < -1/2x + 3

x -4 -2 0 2 4y 5 4 3 2 1

3x - y > 63x - 6 > yy < 3x - 6

x -2 -1 0 1 2y -12 -9 -6 -3 0

y _ 1/2x - 8x -4 -2 0 2 4y -10 -9 -8 -7 -4

Graph all three lines on the same set of axes and shade the area designated by each.

The area designated by the darkest shading is the intersection of this system of equations. Points within this region satisfy all three inequalities.

For Example: Graph the intersection of: and Solution: Set up tables of values;

x -2 -1 0 1 2y -1 -1 -1 -1 -1

x -2 -1 0 1 2y -7 -5 -3 -1 1

Draw the graph and shade the area that satisfies both

In the previous example we can see that a system of two lines can divide a plane into four different areas;

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Unit 3: Matrices

Practice Questions 11: Systems of Inequalities.1. Write the inequalities that define each of the shaded regions in the following graphs. The

equations of the lines are given.

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Unit 3: Matrices2. Plot the graph of the intersection of the following systems of inequalities:

a. b.

c. d.

e. f.

g. h. -3x + y < 9

i. j. How many lines are needed to completely enclose a region of the graph?

Linear Programming

Many situations in life are described by systems of inequalities. Businesses and governments often need to find situations which produce maximum profits or minimum costs. Often the factors which control the profits or costs are inequalities.

In order to find the optimum solution we will need to use the following steps:

1. Identify the variables. Use a “let” statement to define the factors which are in question.

2. Create inequalities to describe the constraints on the system. These are the limiting factors for the variables, they usually be described using terms like “at least”( ) or “the most”( ) as well as less than and greater than and “total”.

3. Create an equation to describe the quantity being optimized. Usually profits or costs. The type of optimization needed is given by expressions like “maximum”, “minimum”, “greatest”, “least”, “highest”, “lowest”, etc.

4. Graph the constraints. Using the techniques learned in the last section create a graph showing the intersection of the constraint inequalities.

5. Find vertices (plural of vertex) of the area of intersection. These points are the solutions to the pairs of lines that outline the inequalities. You can determine the intersections using matrices or methods for solving systems of equations.

6. Substitute the coordinates of the vertices into the optimization equation. The highest value computed is the maximum for that system. The lowest value computed will be the minimum for that system.

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Unit 3: MatricesFor Example: A canoe maker can build a total of 20 canoes in a season. She must build at least

8 cedar strip canoes and at least 3 wood and canvas canoes. The price is $1500 per cedar strip canoe and $1650 per wood and canvas canoe. The expenses are $800 for the cedar strip and $1000 for the wood and canvas.

a. What number of each type of canoes should she make to minimize costs?b. What number of each canoe should she build to maximize the profit? How much profit

will she make?

Solution: 1. Identify the variables: Let x = the number of cedar strip canoes.

Let y = the number of wood and canvas canoes.

2. Create inequalities to describe the constraints on the system. (look for “at least” and “total”)“At least 8 cedar strip”. x ≥ 8“At least 3 wood and canvas”. y ≥ 3“Total of 20 canoes:. x + y ≤ 20

3. Create an equation to describe the quantity being optimized.Part a. Cost = $800x + $1000yPart b. Profit = earnings - cost = ($1500x + $1750y) - ($800x + $1000y)

= $700x + $750y4. Graph:

5. Find the vertices of the graph:

The intersection of x = 8 and y = 3 (8, 3) The intersection of x = 8 and x + y = 20; solve by substitution 8 + y = 20 y = 12 Solution (8, 12) The intersection of y = 3 and x + y = 20; solve by substitution x + 3 = 20 x = 17 Solution (17, 3)

6. Substitute the coordinates of the vertices into the optimization equation(s) to find the maximum or minimum:

Part a. Cost = $800x + $1000yCost= $800(8) + $1000(3) = $9400Cost= $800(8) + $1000(12) = $18400Cost= $800(17) + $1000(3) = $16600

Part b. Profit = $700x + $750yProfit = $700(8) + $750(3) = $7850Profit = $700(8) + $750(12) = $14600Profit = $700(17) + $750(3) = $14150

7.a.The minimum cost occurs when she makes 8 cedar strip and 3 wood and canvas canoes b. The maximum profit occurs when she makes 8 cedar strip and 12 wood and canvas.

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Unit 3: MatricesFor Example: Maximizing the profits on a farm. A mixed farm produces wheat and canola on

1500 hectares. The farmer calculating costs and profits on the farm devised the following list;Canola: Cost per hectare = $180 Earnings per hectare = $210Wheat: Cost per hectare = $130 Earnings per hectare = $170

In order to meet her obligations, the farmer must plant at least 500 hectares of canola and 700 acres of wheat. How much of each must be planted to maximize her profits?

Solution: First, define the variables: Let x = hectares of canolaLet y = hectares of wheat

Next, create inequalities to describe the constraints on the system. (Limiting factors)Minimum amounts of wheat and canola: x ≥ 500, y ≥ 700

Total area: → x + y = 1500Create an equation to describe the quantity being optimized.Profit = P = total earnings - total cost

= (210x + 170y) - (180x + 130y) = 30x + 40y

Graph the constraints. Identify the area of intersection and the vertices.

Optimization occurs at the vertices of the intersection of the constraints. To find them solve the pairs of the equations of the lines that define each point.

x = 500, y = 750 (500, 750) x = 500, x + y = 1500 (500, 1000) y = 750, x + y = 1500 (750, 750)

Calculate the value of optimization (profit) at each vertex:

P = 30x + 40y = 30(500) + 40(750) = 45,000 P = 30x + 40y = 30(500) + 40(1000) = 55,000 P = 30x + 40y = 30(750) + 40(750) = 52,500

The solution is the set of values that produces the maximum profit:

The farmer should plant 500 hectares of canola and 1000 hectares of wheat.

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Unit 3: MatricesPractice Questions 12: Linear Programming

1. For each of the following, use the optimization equation and the graph to find the maximum, and minimum optimization.

a. R = 12x -3y b. W = 2x + 6y

2. A linear programming problem gives (0,0), (0, 25), (15, 0), and (8, 10) as the vertices for a set of constraints. If the Optimization equation is Q = 3x - 2y, find the maximum and minimum values of Q.

3. Find the maximum and minimum values of the function S = 10x + 3y in the region defined by the intersection of x ≤ 10, y ≥ 5 and y ≤ 0.5x + 4.

4. A sports wear company manufactures wind suits and duffel bags. They are able to produce a maximum of 30 duffel bags or half that many windsuits (2x + y = 30) per day and need to produce at least 5 of each in order to meet their obligations. The following table provides the cost and sales price for each item;

Item Cost Sales PriceWind suit $23.00 $75.00Duffel bag $10.00 $20.00

a. Create inequalities to describe the constraints on the system.b. Create an equation to describe the profit.c. Graph the constraints.d. Determine the number of each item needed to maximize profits.

5. An arts coop makes and sells hand crafted drums in two sizes. The large drum needs 1.5 square yards of leather and takes 4 hours to build. The small drum needs 1.5 square yards of leather for each side and takes 2 hours to build. Materials for the large drum cost $35 and it sells for $120. The cost of the small drum is $25 and it sells for $50. If the coop has 80 hours and 45 square yards of leather:

a. What number of each would minimize the costs? (Suppose the business is just starting.)

b. What number of each would maximize the profit? What profit can be expected?

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Unit 3: MatricesUnit 3 Solutions

Practice Questions 1: Page 43

1. a. A3×3, B1×4, C2×4, D3×5. b. A - 9, B - 4, C - 8, D - 15 c. A – [3, 2, 5], B – [13], C – [e, u], D – [0.5, , 1]d. A – [6, 4, 3], B – [7, 11, 13, 17], C – [q, w, e, r], D – [ , , 0.5, 4.28, 0]e. a22 = 0, c21 = t, d32 = , b24 = Ø

2. a. A1×2 b. B1×1 c. C4×1 d. D5×3 e. E3×3 f. F1×3

3. a. x = 5, z = 3 b. x = , y = , z = c. x = 3, y = -8, z = 4 d. x = 3, z =

4. a. , , ,

b. a22 = = 0, c21 = , d32 = = , b24 = = Ø

5. , , , , ,

6.

Practice Questions 2: Page 45

1. a. b. c. d. Not possible. e.

2. a. b. c. d.

3. a. b. c. d. e.

4. a. b.

c.

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Unit 3: MatricesPractice Questions 3: Page 48

1. a. ,

b. c.

2. Network

PS

RQ

3. Matrix: 4. Matrix:

5.

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Unit 3: MatricesPractice Questions 4: Page 511. a. P2×2 b. P3×3 c. not d. P2×3 e. not f. P1×3 g. not h. P1×2 i. P3×2 j. P1×3 k. not l. P3×3

2. a. [11], b. , c. , d. e. Not possible f.

3. a. b. c. d.

e. f. g. S is the identity. h. Q and R multiply to give the

identity.

4. a. , b.

Practice Questions 5: Page 53

1. a. b. c. d. e. f. g.

2. a. b.

c. d.

Practice Questions 6: Page 57

1. a b. c. Can’t be done (A) = 0 d. e. Can’t be done (A) = 0

f. g. h. 2. see question.

Practice Questions 7: Page 58

1. a. b. c. d.

e. f.

Practice Questions 8: Page 61

1. a. (x, y) = (3, 2) b. (x, y) = (-2, -7) c. (x, y) = ( , ) d. (x, y) = (-5, 6) e. (x, y) = ( , ) f. (x, y) = (no answer) g. (x, y) = (no answer) h. (x, y) = (5, 18)

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Unit 3: Matrices

2. , ,

The first truck carries 4 tonnes, the second carries 5 tonnes.Practice Questions 9: Page 64

1. Answers will vary.

2. a. b. c. d.

e. f. no solution g. h.

i. j.

Practice Questions 10: Page 661.

2. a.y > 2(y - 3) b. x + 2 ≤ 2(x - 3)

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Unit 3: Matrices

c. 2x + y ≥ 4 d. y > x - 3

e. 3y - x ≤ x + 2 f. 6x - 4y < 2(6 - y)

3. a. 2x + y < 1 b. x - 2y ≤ 3

Practice Questions 11: Page 68

1. a. x > -4, y ≥ b. y ≥ , 14y > x c. y > -4, 2y - 3x ≤ 14 , y + x ≤ 2d. x > -5, y > -2, x + 3y ≤ 9, 2y + 3x ≤ 2

2. a. x ≥ -2, y ≤ 7 b. y - x < 3, x ≤ 5

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Unit 3: Matricesc. 2x - y > 5, y ≤ -2x + 1 d. 4x + 2y ≤ 5, 3x - y ≥ -2

e. y + 2 > 0, x + 1 < y, 3x + y ≤ 4 f. y - 2x ≤ 3, x - 1 ≤ 0, x + 5y ≥ 7

g. 2x - 5y < 10, 3x + 4y ≤ 12, x≥ 0, y ≥ 0 h. x - 4y < -4, 3x + y ≤ 9, -3x + y < 9

i. 2x - y < 1, x - 3y ≥ -5, 4x - y ≥ -6 j. Three lines are needed to completely enclose.

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Unit 3: MatricesPractice Questions 12: Page 72

1. a. Rmax = 12(30) - 2(60) = 180, Rmin = 12(10) - 2(60) = -60b. Wmax = 2(2.5) + 6(7.5) = 50, Wmin = 2(0) + 6 (0) = 0

2. Qmax = 3(15) -2(0) = 45, Qmin = 3(0) - 2(25) = -50

3. Smax = 10(10) + 3(9) = 127, Smin = 10(2) + 3(5) = 35

4. a. Let x = number of windsuitsLet y = number of duffel bagsx _ 5, y _ 5 2x + y _ 30

b. Profit = x($52) + y($10)

d. Pmax = 10($52) + 10($10) = $620

Maximum profit of $620 occurs when they produce 10 windsuits and 10 duffelbags.

5. Let x = number of large drumsLet y = number of small drums

4x + 2y ≤ 801.5x + 2(0.75)y ≤ 45

Cost = 35x + 25yProfit = 85x + 25y

Costs are minimum if they make 20 large drums, $700.

Profits are maximum if they make 20 large drums, $1350.

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Unit 3: MatricesUnit 3 Review

1. A Matrix is a rectangular array of numbers. The elements or entries of the matrix may be identified by Rows (horizontal) or Columns (vertical).The number of rows number of columns are the dimensions of the matrix.

For example: , A is a 2 by 3 matrix. a23 is the entry in row 2 column 3.

2. The transpose of a matrix is the matrix that results when the rows and columns are interchanged.

For example: ,

3. Matrices of the same dimension may be added by adding the corresponding entries. Subtraction is achieved by adding the additive inverse of each of the entries.

4. Multiplication by a scalar (real number) is achieved by multiplying each entry of the matrix by the scalar. Division by a scalar is best approached as multiplication by the reciprocal of the scalar.

5. Matrix addition and scalar multiplication are Commutative (A + B = B + A, nA = An) and Associative ( A + (B + C) = (A + B) + C, a(bC) = (ab)C ). Scalar multiplication is Distributed over addition a(B + C) = aB + aC

6. Any table of numbers can be written as a matrix. Matrices can also be used to describe networks (pg 47) and geometric figures (Incidence matrix pg. 47).

7. Matrix Multiplication can only be done between two matrices if the number of columns of the first is the same as the number of rows of the second.

Each entry of the product matrix is found by adding the product of the entries of a row of the first matrix with the corresponding entries of a column of the second. For example entry p11 of the product matrix is the sum of the products of row 1 from the first matrix times column 1 of the second.

abcd

wxyz

aw by

abcd

wxyz

aw

bby ax bz

abcd

wxyz

aw by ax bzcw dy

a

bbcd

wxyz

aw by ax bzcw dy cx dz

Matrix multiplication is not commutative. (AB ≠ BA)

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Unit 3: Matrices8. The identity matrix (I)is a square matrix whose entries along the upper left to lower right

diagonal are all ones and the rest are zeros. This matrix has the property that A×A for any square matrix, A.

9. The determinant of a square matrix is a number calculated by multiplying and adding(and subtracting) the diagonals. (pg 52)

For the 2×2 matrix A = , the determinant det(A) = (A) = = ad - bc.

10. The multiplicative inverse of a square matrix is a matrix which, when multiplied by the original matrix produces the identity. If A× B = I then A and B are inverses. A-1 = B and B-1 = A. The inverse for a 2×2 matrix can be found by using the following:

11. The inverse of a matrix can used to solve a matrix equation: If AB = C, then B = A-1C.

12. Matrices can be used to solve systems of equations by:

i) Using the inverse of a matrix;

ii) Using row operations on an Augmented Matrix (pg 61 to 63)There are three Elementary Row Operations:

1.Interchange any row with another row.2.Multiply any row by a constant.3.Replace any row by adding (or subtracting) any other row.

Using these row operations it is possible to reduce an augmented matrix to the form;

Which corresponds to: or

13. Inequalities in two variables may be plotted on the Cartesian plane by graphing the line of the associated equation and shading the appropriate region of the plane. If the inequality (when solved for y) is of the form “y > “ or “y ≥ “ the upper half of the plane should be shaded. If the inequality is of the form “y < “ or “y ≤ “ the lower half of the plane should be shaded.

14. Systems of inequalities can be graphed by shading the area of intersection of the inequalities.

15. Linear Programming (pg 69) is a technique which uses systems of inequalities to describe the constraints on some real system. The Optimum (greatest or least) value can be found by using the coordinates of the vertices of the intersection of the constraints in an Optimization equation.

Unit 3 Review Questions-88-

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Unit 3: Matrices

1. Given the following matrices:

, , , &

a. State the name and dimensions of each using subscripts.b. List the elements in the second column.c. List the elements in the first row.d. Write the transpose matrix for each.e. Write the following elements; , and,

2. Find the values for q, r & s for which the following matrices are equivalent.

a. b.

c. d.

3. Given the matrices:

, , , and

Calculate the following, if possible:

a. Q + R b. R - Qc. R + T d. 3S - Te. Q ÷ 3 f. 2T + QT

g. Solve for the matrix X if X + S = T h. Solve for matrix Y if Q - Y = R

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Unit 3: Matrices4. Create a matrix to illustrate the following;

a. The network below; b. An incidence matrix for the figure;

c. The school canteen made the following sales in January, 273 bags of chips, 116 sandwiches, and 223 chocolate bars. In February their sales were, 189 bags of chips, 153 sandwiches and 187 chocolate bars. The profit on each is 12 ¢ per bag of chips, 23¢ per sandwich and 9¢ per chocolate bar. Write a matrix to describe the sales, another matrix to describe the prices and use matrix multiplication to calculate the monthly profit.

5. Given the following matrices: Q13, W31, E23, R22, T33, & Y34: Indicate if each of the following multiplications are possible. If they are possible give the dimensions of the product (Prc).

a. QW b. WQc. ER d. REe. T2 f. ETRg. Y2 h. WQYi. E2 j. RT

6. Multiply the following, if possible:

a. b.

c. d.

e. f.

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Unit 3: Matrices7. Use the inverse to solve the following;

a. b.

c. 5x + 3y = 5 d. 7x + 6y = 9 2x + y = 1 -4x - 3y = -3

8. Solve the following systems using row operations:

a. 3x + y = 7 b. 4x + 5y = 82x - 5y = -1 6x - 7y = 0

c. 3x +2y + z = 10 d. 4x + y +3z = 11x - y - 2z = 6 -3x + 2y - z = 07x + 8y - 6z = 5 2x - 3y + 2z = 10

9. Find the determinant for each of the following;

a. b.

10. Graph each of the following systems of inequalities:

a. y ≤ 3 b. x > -3

c. x - y ≤ -6 d. 2x - 3y > 12

e. f.

g. h.

11. A recent math contest for high school math students had two categories of questions. Questions in section A were allowed 4 minutes to solve and were worth 6 points each. Those from section B took 6 minutes each and were worth 10 points each. Each student is allowed to answer 12 questions within 60 minutes. How many of each question should be answered in order to obtain the best possible score. (I assume that all answers will be correct, of course.)

a. Identify the variables for this system.b. Create inequalities to describe the constraints.c. Create an optimization equation (total points).d. Graph the system.e. Find the optimum solution.

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Unit 3: MatricesUnit 3 Review Questions Solutions

1. a. L3×3, U3×1, G2×5, & N1×4 b.L: l12 = 3,l22 = 6,l32 = -3, U: Ø, G:g12 = r, g22 = a, N: n12 = c. L: l11 = 5, l12 = 3, l13 = -7, U: u11 = 5, G: g11 = a, g12 = r, g13 = g, g14 = l, g15 = e

N: n11 = , n12 = , n13 = , n14 = 1

d. , , , &

e. l23 = 9, u13 = 7,

2. a. q = 5, s = -3 b. q = 12, r = -5, s = 3 c. q = 0, r = 5, s = -7 d. q = , s = 1/4, r = -3

3. a. b. c. Can’t do d. e.

f. g. h.

4 a. b.

c.

5. a. yes P1×1 b. yes P3×3 c. not possible d.yes P2×3 e. yes P3×3 f. not possibleg. not possible h. yes P3×4 i. not possible j. not possible

6. a. ,b. ,c. ,d. ,e. [29] f.

7. a. b. c. d.

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Unit 3: Matrices

8. a. b. c. d.

9. a. (A) = 7 b. (A) = 11

10. a. y ≤ 3 b. x > -3

c. x - y ≤ -6 d. 2x - 3y > 12

e. y < 3, 4y - 8x ≥ 10 f. x > -5, 2x - 3y < 6

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Unit 3: Matrices

g. x + 3 > 0, x + y ≤ 6, 2x - y < 6 h. y ≥ 4x - 3, -3x + y ≤ 4, 2x + y ≤ 3

11. a. Let x = 4 min questions Let y = 6 min questions

b. 4x + 6y ≤ 60x ≤ 12y ≤ 12x & y ≥ 0

c. Total = 6x + 10y

e. Intersections of lines:(0, 10) Total = 100(12, 0) Total = 72(12, 2) Total = 92

d. The graph

The best possible score is 100 by doing 10 of the six minute/ 10 point questions.

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Unit 3: Matrices

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