needs work need to add –hw quizzes chapter 13 matrices and determinants

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• Need to add – HW Quizzes

Chapter 13

Matrices and Determinants

13.1 Matrices and Systems of Equations

row

nmrows

mnmmm

n

n

n

aaaa

a

a

a

aaa

aaa

aaa

A

321

3

2

1

333231

232221

131211

A matrix is a rectangular array of numbers. We subscript entries to tell

their location in the array

Matrices are

identified by their

size.

14

51 20513

3

1

6

2

0974

9852

7531

4212

44

A matrix that has the same number of rows as columns is called a square matrix.

main diagonal

44434241

34333231

24232221

14131211

aaaa

aaaa

aaaa

aaaa

A

174

242

3523

zyx

zyx

zyx

741

412

523

ACoefficient matrix

If you have a system of equations and just pick off the coefficients and put them in a matrix it is called a coefficient matrix.

174

242

3523

zyx

zyx

zyx If you take the coefficient matrix and then add a last column with the constants, it is called the augmented matrix. Often the constants are separated with a line.

1741

2412

3523#AAugmented matrix

We are going to work with our augmented matrix to get it in a form that will tell us the solutions to the system of equations. The three things above are the only things we can do to the matrix but we can do them together (i.e. we can multiply a row by something and add it to another row).

Operations that can be performed without altering the solution set of a linear system

1. Interchange any two rows

2. Multiply every element in a row by a nonzero constant

3. Add elements of one row to corresponding elements of another row

#100

##10

###1

After we get the matrix to look like our goal, we put the variables back in and use back substitution to get the solutions.

We use elementary row operations to make the matrix look like the one below. The # signs just mean there can be any number here---we don’t care what.

#100

##10

###1

Suse row operations to obtain row echelon form:

1762

353

12

zyx

zyx

zyx

1762

3153

1121

The augmented matrix

Work on this column first. Get the 1 and then use it as a “tool” to get zeros below it with row operations.

We already have the 1 where we need it.

We’ll take row 1 and multiply it by -3 and add to row 2 to get a 0. The notation for this step is -3r1 + r2 we write it by the row we replace in the matrix (see next screen).

1762

0210

1121

1762

3153

1121

1762

0210

1121

-3r1 + r2

-3r1 -3 -6 -3 -3 + r2 3 5 1 3

0 -1 -2 0

Now we’ll use -2 times row 1 added to row 3 to get a 0 there.

-2r1 + r3

1520

0210

1121

-2r1 -2 -4 -2 -2 + r3 2 6 7 1

0 2 5 -1

Now our first column is like our goal.

1520

0210

1121

- 1r2

-2r2 0 -2 -4 0 + r3 0 2 5 -1

0 0 1 -1

-2r2 + r3

1520

0210

1121

We’ll use row 2 with the 1 as a tool to get a 0 below it by multiplying it by -2 and adding to row 3

#100

##10

###1Now we’ll move to the second column and do row operations to get it to look like our goal.

We need a 1 in the second row second column so we’ll multiply row 2 by a -1

1100

0210

1121

the second column is like we need it now

#100

##10

###1

Now we’ll move to the third column and we see for our goal we just need a 1 in the third row third column and we have it so we’ve achieved the goal and it’s time for back substitution. We put the variables and = signs back in.

1100

0210

1121Substitute -1 in for z in second equation to find y

1

02

12

z

zy

zyxx

colu

mn

y co

lum

n

z co

lum

n

equal signs 012 y

2y

Substitute -1 in for z and 2 for y in first equation to find x.

1122 x

2x

Solution is: (-2, 2, -1)

Solution is: (-2, 2, -1)

1762

353

12

zyx

zyx

zyx

This is the only (x, y, z) that make ALL THREE equations true. Let’s check it.

1172622

312523

11222

These are all true.

Geometrically this means we have three planes that intersect at a point, a unique solution.

#100

#010

#001

This method requires no back substitution. When you put the variables back in, you have the solutions.

To obtain reduced row echelon form, you continue to do more row operations to obtain the goal below.

Let’s try this method on the problem we just did. We take the matrix we ended up with when doing row echelon form:

1762

353

12

zyx

zyx

zyx

Let’s get the 0 we need in the second column by using the second row as a tool.

-2r2+r1

Now we’ll use row 3 as a tool to work on the third column to get zeros above the 1.

#100

#010

#001

1100

0210

1121

1100

0210

1301

-2r3+r2

3r3+r1

1100

0210

2001

1100

2010

2001

Notice when we put the variables and = signs back in we have the solution

1,2,2 zyx

Let’s try another one:

1237

0432

6223

zyx

zyx

zyxThe augmented matrix:

1237

0432

6223

If we subtract the second row from the first we’ll get the 1 we need for the first column.

r1-r2

We’ll now use row 1 as our tool to get 0’s below it.

-2r1+r2

1237

0432

6211

1237

12850

6211

-7r1+r3

4316100

12850

6211We have the first column like our goal. On the next screen we’ll work on the next column.

#100

##10

###1

1237

0432

6223

zyx

zyx

zyx

If we multiply the second row by a -1/5 we’ll get the one we need in the second column.

We’ll now use row 2 as our tool to get 0’s below it.

-1/5r2

10r2+r3

4316100

12850

6211

Wait! If you put variables and = signs back in the bottom equation is 0 = -19 a false statement!

43161005

12

5

810

6211

190005

12

5

810

6211

#100

##10

###1INCONSISTENT - NO SOLUTION

534

132

465

zyx

zyx

zyx

5134

1132

4165 One more:

#100

##10

###1

r1-r3

-2r1+r2

-4r1+r3

5134

1132

1231

9990

3330

1231

1/3r2

9990

1110

1231

-9r2+r3

0000

1110

1231

Oops---last row ended up all zeros. Put variables and = signs back in and get 0 = 0 which is true. This is the dependent case. We’ll figure out solutions on next slide.

5134

1132

1231

0000

1110

1231

x y z

zz

zy

zx

1

2

zz

zy

zx

1

2

0000

1110

2101

Let’s go one step further and get a 0 above the 1 in the second column

3r2+r1

Infinitely many solutions where z is any real number

No restriction on z

put variables back in

solve for x & y

zz

zy

zx

1

2

Infinitely many solutions where z is any real number

534

132

465

zyx

zyx

zyx

What this means is that you can choose any real number for z and put it in to get the x and y that go with it and these will solve the equation. You will get as many solutions as there are values of z to put in (infinitely many).

Let’s try z = 1. Then y = 2 and x = 3

512334

112332

412635

works in all 3

Let’s try z = 0. Then y = 1 and x = 2

501324

101322

401625

The solution can be written: (z + 2, z + 1, z)

HW #13.1Pg 572 1-4, 6-7, 9-11, 15

HW Quiz 13.1Wednesday, April 19, 2023

Row 1, 3, 5

1. 4

2. 6

3. 10

4. 15

Row 2, 4

1. 2

2. 4

3. 6

4. 10

Chapter 13Matrices and Determinants

Section 13.2

Addition and Subtraction of Matrices

To Add and Subtract matrices

To find the additive inverse of a matrix

To compare matrices they must have the same dimensions and have the same entries in the same

positions

You can only add or subtract matrices when they have exactly the same dimensions

4. The operation is not possible

The additive inverse of a matrix can be obtained by replacing each element by its additive inverse.

Finding the Additive Inverse of a Matrix

Find the additive inverse of the matrix

12 2 15

16 0 9

9 13 3

13 3 2 10

Find the Additive Inverse of each Matrix

2 5

1 3

5 0

2 1

4 3

4 3 2

1 5 4

2 7 6

3 0 10 8

Subtracting by finding the Additive Inverse

Subtract by finding the Additive Inverse

2 1

4 2

2 3

1 3

Exercises for Example 4

Subtract by finding the Additive Inverse

HW # 13.2

Pg 575 1-32


Section 13.3

Cramer’s Rule

Objective: Evaluate a 2 x 2 Determinant

A B

Your Turn Hidden

Objective: Solve a system of 2 equations and 2 variables using Cramer’s Rule

Objective: Evaluate a 3 x 3 Determinant

Your Turn Hidden

C

Objective: Solve a system of 3 equations and 3 variables using Cramer’s Rule

ED

Your Turn Hidden

HW #13.3Pg 580 1-33 odd, 34-42


Section 13.4

Multiplying Matrices

4 X 3 3 X 5 4 X 5

A B AB

MULTIPLYING TWO MATRICES

4 rows4 rows

3 columns3 columns

3 rows3 rows

5 columns5 columns

4 X 3 3 X 5 4 X 5


4 rows4 rows

5 columns5 columns

4 rows4 rows

5 columns5 columns

A B AB

If A is a 4 X 3 matrix and B is a 3 X 5 matrix, then the product AB is a 4 X 5 matrix.


m X n n X p m X p

A B AB


m rowsm rows

n columnsn columns

n rowsn rows

p columnsp columns

m X n n X p m X p


m rowsm rows

p columnsp columns

m rowsm rows

p columnsp columns

A B AB

If A is an m X n matrix and B is an n X p matrix, then the product AB is an m X p matrix.


Finding the Product of Two Matrices

– 2 3 1 – 4 6 0

– 1 3– 2 4

Find AB if A = and B =

Use a similar procedure to write the other entries of the product.

Because A is a 3 X 2 matrix and B is a 2 X 2 matrix, the product AB is defined and is a 3 X 2 matrix.

To write the entry in the first row and first column of AB, multiply corresponding entries in the first row of A and the first column of B. Then add.

SOLUTION

(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)

(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)

(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)

3 X 2 2 X 2 3 X 2

A B AB

– 2 3

1 – 4

6 0

– 1 3

– 2 4


3 X 2 2 X 2 3 X 2

A B AB


(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)

(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)

(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)

– 2 3

1 – 4

6 0

– 1 3

– 2 4

3 X 2 2 X 2 3 X 2

A B AB


– 4 6

7 – 13

– 6 18

(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)

(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)

(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)

3 x 2 2 x 2

AB will be 3 x 2

2 x 2 2 x 2

AB will be 2 x 2

2 x 2 2 x 2

BA will be 2 x 2

Properties of Matrix Arithmetic

• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Commutative Property of Addition

• A + B = B + A

– Associative Property• A + (B + C) = (A + B) + C

• A(BC) = (AB)C


• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Additive Identity

• There exists a unique matrix O such that A + 0 = 0 + A = A

– Additive Inverse• There Exists a unique matrix –A such that

A + (-A) = -A + A = 0


• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Distributive Property

• A(B + C) = AB + AC


• For any real numbers k and mk(A + B) = kA + kB(k + m)A = kA + mA(km)A = k(mA)

HW #13.4

Pg 587-588 1-39 Odd, 40-46

HW Quiz 13.4Wednesday, April 19, 2023

Row 1, 3, 5

Write the answer to

1. 9

2. 15

3. 17

4. 39

Row 2, 4, 6

Write the answer to

1. 9

2. 15

3. 17

4. 39

1 34 6 14 71.3 1 3 2

14 7

4 6 1 02.

3 1 0 1

3 1 28 8 83 1 01 3 6

3. 1 1 28 8 8

1 1 1 2 2 48 8 8

1 0

0 1

4 6

3 1

1 0 0

0 1 0

0 0 1


Section 13.5

Inverses of Matrices

To write a matrix equation equivalent to a system of matrices

To determine when two matrices are multiplicative inverses and find the multiplicative inverse of a 2 x 2 matrix

Two n x n matrices are inverses of each other if their product (in both orders) is the n x n identity matrix.

AA-1 = A-1A = 1

To determine if two matrices A and B are inverses of each other you need to make sure AB = BA = I

Determine if A and B are multiplicative inverses of each other

3 4

2 6A

3 25 51 35 10

B

No

2 3

3 6C

Determine if C and D are multiplicative inverses of each other

2 1

21

3

D

Yes

Tell whether the matrices are multiplicative inverses of each other.

Computing an Inverse Matrix 2 x 2

Let’s Prove it!

Use the shortcut

Find the inverse of the given matrix

2 1 1

1 2 3

4 1 2

7 1 1

15 5 152 1

03 33 2 1

5 5 5

HW #13.5

Pg 591-592 1-19, 21-25 Odd

13.6

Inverses and Systems

Writing Linear Systems as Matrix Equations

Consider the system

Let A = Let X = Let B =

Write the equation AX = B using the above matrices

Coefficient Matrix

Matrix of Variables

Matrix of Constants

For a linear system of equations written as a matrix equation the matrix A is the coefficient matrix of the system, X is the matrix of variables, and B is the matrix of constants.

Write the system of linear equations as a matrix equation

4. 2 3 4

2 1

4 1

x y z

x y z

x y z

2 1 1

3 2 0

x

y

3 4 4

4 5 7

x

y

6 5 3

3 2 3

x

y

1 2 3 4

2 1 1 1

4 1 1 1

x

y

z

SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B.

Solve system of linear equations.

4. 2 3 4

2 1

4 1

x y z

x y z

x y z

2 1 1

3 2 0

x

y

3 4 4

4 5 7

x

y

6 5 3

3 2 3

x

y

1 2 3 4

2 1 1 1

4 1 1 1

x

y

z

HW 13.6Pg 596-597 1-19 Odd

13-7 Using Matrices

Application

Two stores sell the exact same brand and style of a dresser, a night stand, and a bookcase. Matrix A gives the retail prices (in dollars) for the items. Matrix B gives the number of each item sold at each store in one month.

Calculate AB and interpret the entries of AB

Your Turn

HW #13.7Pg 600-601 1-7

Chapter 13 Test Review

Part 1

Helpful Hints

• What does it mean when the determinant of a matrix is 0– In terms of a system of Linear Equations– In Terms of a Matrix

• Rules of Multiplication/Addition/Equality• Scalar Multiplication• Shortcut for finding the inverse of a 2 x 2

matrix

Evaluate the determinant of the matrix.

Use Cramer’s rule to solve the system of equations.

Use Augmented Matrices to solve the system of equations.

Use the matrix equation AX = B to solve the system.

When does a matrix fail to have an inverse?

Study all the challenge problems in the book

Know how to multiply and add matrices

Chapter 13 Test Review

Part 2

This test will have only 1 part

HW #R-13Pg 609 1-14

needs work need to add –hw quizzes chapter 13 matrices and determinants

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