Calculus for Life Sciences · 2013-01-22 · Calculus for Life Sciences MAT 1332 C Winter 2010 Jing...

Calculus for Life SciencesMAT 1332 C Winter 2010

Jing Li

Department of Mathematics and StatisticsUniversity of Ottawa

March 8, 2010

Jing Li (UofO) MAT 1332 C March 8, 2010 1 / 39

Outline

1 Midterm 1 Grade Statistics

2 Feedback Questionaire

3 Linear Algebra I: Solving Linear Systems of Equations

4 Linear Algebra II: Vectors and MatricesDefinitionOperations

Basic Matrix OperationsMatrix-Vector MultiplicationMatrix-Matrix Multiplication

Midterm 1 Grade Statistics

Table: Midterm 1 Grade Statistics

Count: 169Average: 18.8Median: 19.5Maximum: 29.5Minimun: 2.0Standard Deviation: 5.91

Midterm 1 Grade Statistics

Feedback Questionaire

Aspects that help learning

Algorithms

Help available outside class

Examples in class

Assignments

Math help center

Review sessions

Suggestions:

Better English

Better classroom

Better blackboard presentation (notes )

Slow down (especially in the end of class)

Practice midterms

Better control of the class (too noisy)

Better organization and better introduction

More explanation

Application in real lives

DGDs should focus on similar problems for midterms and assignments.

get class more involved in the lecture.

More examples (difficult ones, close to tests and assignments, less parameters)

Better TA picking

do assignment questions in class

more office hours

Changes that can be implementedget help from me, TA, help center.NOTEs (online)Better lecture presentation (organization, introduction, examples,explanation, application)Slow down speedBetter control of the class (quite, get class more involved inlectures)Better questions in DGDs. (do hard assignment questions inDGDs)

Linear Algebra I: Solving Linear Systems of Equations

Reduced row-echelon form

Definition

leading entry, row-echelon form, and reduced row-echelon form

The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.

A matrix is in row-echelon form if the following three rules are true

(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in

any row above it.(3) All entires in the column below a leading entry are zero.

A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition

(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.

Definition

(1) Rows of zeros are blew any nonzero row.

(2) The leading entry of any row is to the right of any leading entry inany row above it.

(3) All entires in the column below a leading entry are zero.

Definition

any row above it.

(3) All entires in the column below a leading entry are zero.

Definition

(4) Each leading entry is 1.

(5) All entries in the column above a leading entry are zero.

Definition

Example 5.

Categorize the following systems according to the above definition:241 2 00 3 10 0 2

242 0 2 40 1 1 00 0 0 7

241 0 20 1 10 0 0

241 0 2 00 1 −3 00 0 0 1

242 0 21 1 10 3 0

35Solution:

row-echelon form but not reduced: (1), (2)

reduced row-echelon form: (3), (4)

neither row-echelon nor reduced row-echelon form: (5)

Example 5.

242 0 2 40 1 1 00 0 0 7

241 0 20 1 10 0 0

241 0 2 00 1 −3 00 0 0 1

242 0 21 1 10 3 0

35Solution:

Example 5.

242 0 2 40 1 1 00 0 0 7

241 0 20 1 10 0 0

241 0 2 00 1 −3 00 0 0 1

242 0 21 1 10 3 0

35Solution:

Example 6: Exactly One Solution

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15

−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 5

0 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50

− 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2

− 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 |

− 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→

24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −10

0 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100

0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0

5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 |

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

(−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 5

0 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50

2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2

3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 |

100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 10

0 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100

0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 |

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

−R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→

24 1 1 0 | 10 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24

1 1 0 | 10 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1

1 0 | 10 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1

0 | 10 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 |

10 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 |

− 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 1

0 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10

1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1

0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 |

− 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

1 0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

0 0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0

0 | 20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 |

20 1 0 | −10 0 1 | 4

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).

8<:x1 + x2 + x3 = 52x1 − x3 = 0

x2 + 4x3 = 15

35−2R1 + R2−−−−−−−→

24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15

35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20

35 (−1)R2;15 R3

−−−−−−−→

24 1 1 1 | 50 2 3 | 100 0 1 | 4

35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1

0 2 0 | − 20 0 1 | 4

−−→

24 1 1 0 | 10 1 0 | − 10 0 1 | 4

35−R2 + R1−−−−−−→

24 1 0 0 | 20 1 0 | −10 0 1 | 4

Example 7: No Solution

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

−2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.

This is clearly impossible.

This means that this system does not have a solution.

Hence, this system is inconsistent.

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 3

0 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30

− 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2

1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 |

− 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 4

0 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40

− 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2

1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 |

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→

24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −4

0 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40

0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 |

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1

Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1

35 −2R1 + R2

−R1 + R3−−−−−−−−−→

24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2

35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2

Example 8: Infinitely Many Solutions

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

−R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.

This equation is a correct statement, and is true for all values of x1, x2, x3.

The second equation is x2 + 2x3 = 2.

We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .

The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .

Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 4

0 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40

1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1

2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 |

20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 2

0 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20

0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 |

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

This equation is a correct statement,

and is true for all values of x1, x2, x3.

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

We introduce a free variable by denoting x3 = t ,

then x2 = 2− 2x3 = 2− 2t .

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8

Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8

35 −R1 + R2

−2R1 + R3−−−−−−−−−→

24 1 −3 1 | 40 1 2 | 20 0 0 | 0

Example 9: Underdetermined System: A System has fewer equations than variables

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2

3 x3 − 13 = 2

3 t − 13 .

We plug these back to the first equation to solve for x1:

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

Hence, the solution set is

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 1

0 − 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10

− 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 |

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

–This time, the last row reads −3x2 + 2x3 = 1.

Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.

Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.

We set x3 = t as a free variable, then we get x2 = 23 x3 − 1

3 = 23 t − 1

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2

Solution: »2 2 −1 | 12 −1 1 | 2

–−R1 + R2−−−−−−→

»2 2 −1 | 10 − 3 2 | 1

3 x3 − 13 = 2

3 t − 13 .

x1 = −x2 +12

x3 +12

= −(23

t − 13

= − t6

{(x1, x2, x3) : x1 = − t6

, x2 =23

t − 13

, x3 = t , t ∈ R}

Example 10: Overdetermined System: a system has more equations than variables

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

4 from the first row, x1 = 52 from the second row.

Since 14 6=

52 , there cannot be a solution.

The system is inconsistent and has no solution.

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 2

0 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20

− 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 |

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 2

0 1 | − 12

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20

1 | − 12

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 |

− 12

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 |

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 |

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

x1 = 14 from the first row, x1 = 5

2 from the second row.

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3

Solution:24 2 −1 | 11 1 | 21 −1 | 3

35−R2 + R3−−−−−−→

24 2 −1 | 11 1 | 20 − 2 | 1

35−12

−−−−→

24 2 −1 | 11 1 | 20 1 | − 1

35R3 + R1

−R3 + R2−−−−−−−−→

24 2 0 | 12

1 0 | 52

0 1 | − 12

−−→

24 1 0 | 14

1 0 | 52

0 1 | − 12

35x1 = 1

Since 14 6=

Example 11: Problem with parameters

x1 + hx2 = −3−2x1 + 4x2 = 6

have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »

1 h | −3−2 4 | 6

2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

The last row gives the equation: (2h + 4)x2 = 0.

If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.

Hence,

if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −3

0 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30

2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 |

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .

If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

if h 6= −2, then there is a unique solution.

if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.

There is so such of value of h so that the system has no solution.

x1 + hx2 = −3−2x1 + 4x2 = 6

1 h | −3−2 4 | 6

–2R1 + R2−−−−−−→

»1 h | −30 2h + 4 | 0

Hence,

Linear Algebra II: Vectors and Matrices Definition

Vectors and Matrices

Definition

A matrix is a rectangular array of numbers A =

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.

If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].

If m = n, then A is a square matrix. eg. a 3× 3 square matrix:

24−1 3 00 1 −15 4 3

35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.

For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

For the above example,

the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

For the above example, the diagonal elements are −1, 1, 3,

and the trace istr(A) = −1 + 1 + 3 = 3.

Definition

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 .

24−1 3 00 1 −15 4 3

A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].

An m × 1 matrix is called a column vector:

26664b1

37775. eg. a 3× 1 column vector:

Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

375, the size of this matrix is usually clear from the

context.

the identity matrix:

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

37775, the size of this matrix is usually clear

from the context.

26664b1

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

context.

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

from the context.

26664b1

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

context.

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

from the context.

26664b1

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

context.

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

from the context.

26664b1

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

context.

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

from the context.

26664b1

the zero matrix:

2640 · · · 0...

. . ....

0 · · · 0

context.

266641 0 · · · 00 1 · · · 0...

.... . .

...0 0 · · · 1

from the context.Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39

Linear Algebra II: Vectors and Matrices Operations

Matrix Addition and Scalar Multiplication

Definition

Matrix AdditionSuppose that A = [aij ] and B = [bij ] are two m × n matrices. Then

C = A + B

is an m × n matrix with entries

cij = aij + bij , for 1 ≤ i ≤ m, 1 ≤ j ≤ n.

Definition

Scalar MultiplicationSuppose that A = [aij ] is an m × n matrix and k is a number. Then

kA = [kaij ]

for 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Matrix Addition and Scalar Multiplication

Definition

Matrix AdditionSuppose that A = [aij ] and B = [bij ] are two m × n matrices. Then

C = A + B

is an m × n matrix with entries

cij = aij + bij , for 1 ≤ i ≤ m, 1 ≤ j ≤ n.

Definition

Scalar MultiplicationSuppose that A = [aij ] is an m × n matrix and k is a number. Then

kA = [kaij ]

for 1 ≤ i ≤ m and 1 ≤ j ≤ n.

What are the sizes of C = A + B and kA?

C is an m × n matrix.kA is the same size as A. (m × n matrix)

In the field of Linear Algebra, a number is often called a scalar.

Example 1: Consider the following two 2× 3-matrices,

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

Find A + B, A− B, 5A, 3A + 2B.

Solution:

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

Find A + B, A− B, 5A, 3A + 2B.Solution:

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10

123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 12

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6

− 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6

− 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 6

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

5 10 1520 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

10 1520 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10

1520 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

3 6 912 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

6 912 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

912 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

14 16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

16 18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16

18− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18

− 2 − 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2

− 4 − 6

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4

»17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

17 22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

22 2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22

2710 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 27

10 11 12

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11

»1 2 34 5 6

–, B =

»7 8 9−1 −2 −3

1 A + B =

»1 2 34 5 6

»7 8 9−1 −2 −3

»8 10 123 3 3

–2 A− B =

»1 2 34 5 6

–−»

7 8 9−1 −2 −3

»− 6 − 6 − 65 7 9

–3 5A =

»5 10 15

20 25 30

–4 3A + 2B =

»3 6 9

12 15 18

»14 16 18− 2 − 4 − 6

»17 22 2710 11 12

The Transpose of a Matrix

Definition

Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries

AT = [aji ]

which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.

Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

If AT = A, then A is symmetric. e.g. A =

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? –

AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? – AT is n ×m if A is m × n.

What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? –

a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.

What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? –

a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.

Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

If AT = A, then A is symmetric.

e.g. A =

24 1 2 32 4 63 6 5

35, then AT = A.

Definition

AT = [aji ]

24 1 2 32 4 63 6 5

35, then AT = A.

Example 2: Consider

»1 2 34 5 6

–, B =

˜, C =

Find AT , BT , CT .Solution:

24 1 42 53 6

35BT =

–CT =

Example 2: Consider

»1 2 34 5 6

–, B =

˜, C =

24 1 42 53 6

–CT =

Example 2: Consider

»1 2 34 5 6

–, B =

˜, C =

24 1 42 53 6

35BT =

CT =ˆ

Example 2: Consider

»1 2 34 5 6

–, B =

˜, C =

24 1 42 53 6

35BT =

–CT =

Matrix-Vector Multiplication

Definition

If the matrix A has n columns and the column vector x has n rows, then the productAx is defined as follows:

26664a11x1 + a12x2 + · · ·+ a1nxn

a21x1 + a22x2 + · · ·+ a2nxn...

am1x1 + am2x2 + · · ·+ amnxn

37775where,

26664a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

37775 , x =

26664x1

37775 .

How to remember this definition?

simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2

...am1x1 + am2x2 + · · ·+ amnxn = bm

A: coefficient matrix of the linear system.

x1x2...

, unknown variable column vector. b =

b1b2...

What is the size of Ax? –m × 1 column vector.

...am1x1 + am2x2 + · · ·+ amnxn = bm

x1x2...

b1b2...

...am1x1 + am2x2 + · · ·+ amnxn = bm

x1x2...

b1b2...

...am1x1 + am2x2 + · · ·+ amnxn = bm

x1x2...

b1b2...

What is the size of Ax? –

m × 1 column vector.

...am1x1 + am2x2 + · · ·+ amnxn = bm

x1x2...

b1b2...

Example 3: Calculate

»1 2 34 5 6

–24 123

1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1

+ 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2

+ 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 3

4 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1

+ 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2

+ 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)

4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 2

4 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)

4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)

3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

− 3− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3

− 7− 9

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7

»1 2 34 5 6

–24 123

»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3

»1432

»1 2 3 74 5 6 8

–2664123−1

3775 =

»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)

1 2 3˜ 24 1−23

35 =ˆ

1 · 1 + 2 · (−2) + 3 · 3˜

»1 24 5

– »−12

»1 · (−1) + 2 · 24 · (−1) + 5 · 2

24 1 24 53 6

35» 1−2

24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)

24 − 3− 7− 9

Matrix-Matrix Multiplication

Definition

Suppose that A = [aij ] is an m × n matrix and B = [bij ] is an n × k matrix. Then

C = AB

is an m × k matrix with

cij = ai1b1j + ai2b2j + · · ·+ ainbnj =nX

ailblj

for 1 ≤ i ≤ m and 1 ≤ j ≤ k.

How to understand this definition by using the definition of matrix-vectormultiplication?

We may think of each column of B as a column vector of length n.We know how to multiply each of these with the matrix A and thenwe put the resulting vectors into one matrix.

26664a11 · · · a1n

a21 · · · a2n...

...am1 · · · amn

37775 , B =

26664b11 · · · b1k

b21 · · · b2k...

...bn1 · · · bnk

37775 =ˆ

B1 | B2 | · · · | Bk˜

where, B1 =

264 b11...

375, B2 =

264 b12...

375, . . . , B1 =

264 b1k...

Then AB =ˆ

AB1 | AB2 | · · · | ABk˜

What is the size of matrix C, the product of A and B?

(m × n) ∗ (n× k) = (m× k).

How to understand this definition by using the definition of matrix-vectormultiplication?

We may think of each column of B as a column vector of length n.We know how to multiply each of these with the matrix A and thenwe put the resulting vectors into one matrix.

26664a11 · · · a1n

a21 · · · a2n...

...am1 · · · amn

37775 , B =

26664b11 · · · b1k

b21 · · · b2k...

...bn1 · · · bnk

37775 =ˆ

B1 | B2 | · · · | Bk˜

where, B1 =

264 b11...

375, B2 =

264 b12...

375, . . . , B1 =

264 b1k...

Then AB =ˆ

AB1 | AB2 | · · · | ABk˜

What is the size of matrix C, the product of A and B?

(m × n) ∗ (n× k) = (m× k).

Example 4: Consider the following matrices

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

–Then, well-defined colored in blue, undefined colored in red.

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

»1 23 4

–, B =

»1 2 3−1 0 4

–, C =

24 2 11 −1−1 0

35 , D =

»−1 −20 1

»1 23 4

– »1 2 3−1 0 4

»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4

»−1 2 9−1 6 25

–BA =

»1 2 3−1 0 4

– »1 23 4

–(2× 3) ∗ (2× 2) wrong.

BT A =

24 1 −12 03 4

35» 1 23 4

24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

24 −2 −22 415 22

35Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

24 2 11 −1−1 0

35» 1 23 4

24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4

24 5 8−2 −2−1 −2

»1 23 4

– »2 1 −11 −1 0

»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0

»4 −1 −110 −1 −3

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

Example 4 cont’d

A2 = AA =

»1 23 4

– »1 23 4

»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4

»7 10

»1 23 4

– »−1 −20 1

»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1

»−1 0−3 −2

»−1 −20 1

– »1 23 4

»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4

0 · 1 + 1 · 3 0 · 2 + 1 · 4

»−7 −103 4

»−1 0−3 −2

–, DA =

»−7 −103 4

–,−→ AD 6= DA.

The order of the product matters!

Matrix multiplication is NOT commutative, even if both products are defined.

»−1 0−3 −2

–, DA =

»−7 −103 4

–,−→ AD 6= DA.

»−1 0−3 −2

–, DA =

»−7 −103 4

–,−→ AD 6= DA.

Multiplication of the identity matrix

For real number arithmetic, we know that 1 · a = a · 1 = a.

The identity matrix gives the same result in matrix multiplication

If A is n ×m matrix, then we have

InA = AIm = A

Note: we really do need different identity matrices on each side of A that will dependupon the size of A.

InA = AIm = A

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

»1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

Hence,I3A = AI2 = A.

Example 5:

24 10 0−3 8−1 11

24 1 0 00 1 00 0 1

35 24 10 0−3 8−1 11

24 10 0−3 8−1 11

35AI2 =

24 10 0−3 8−1 11

35 » 1 00 1

24 10 0−3 8−1 11

35Hence,

I3A = AI2 = A.

Operations of the zero matrix:

The zero matrix (denoted by 0 for a general matrix and 0 for a column/row vector) willtake the place of the number 0 in most of the matrix arithmetic.

Zero matrix properties: In the following properties, A is a matrix and 0 is the zeromatrix sized appropriately for the indicated operation to be valid.

1 A + 0 = 0 + A, (A is m × n, 0 is m × n)2 A− A = 0, (A is m × n, 0 is m × n)3 0− A = −A, (A is m × n, 0 is m × n)4 0A = 0, (A is m × n, the first 0 is k ×m, the second 0 is k × n)5 A0 = 0, (A is m × n, the first 0 is n × k , the second 0 is m × k )

The zero matrix (denoted by 0 for a general matrix and 0 for a column/row vector) willtake the place of the number 0 in most of the matrix arithmetic.Zero matrix properties: In the following properties, A is a matrix and 0 is the zeromatrix sized appropriately for the indicated operation to be valid.

1 A + 0 = 0 + A, (A is m × n, 0 is m × n)2 A− A = 0, (A is m × n, 0 is m × n)3 0− A = −A, (A is m × n, 0 is m × n)4 0A = 0, (A is m × n, the first 0 is k ×m, the second 0 is k × n)5 A0 = 0, (A is m × n, the first 0 is n × k , the second 0 is m × k )

Now, in real number arithmetic we know that

if ab = ac, and a 6= 0, then b = c. (cancellation law)

if ab = 0 then a = 0 and/or b = 0. (the zero factor property)

BUT, we see the following two examples:

Now, in real number arithmetic we know that

if ab = ac, and a 6= 0, then b = c. (cancellation law)

if ab = 0 then a = 0 and/or b = 0. (the zero factor property)

BUT, we see the following two examples:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

calculate AB and AC:

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

Note:AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

AB = AC , (A 6= 0, B 6= C).

So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.

The cancellation law will not be valid in general for matrix multiplication.

Example 6:

Consider

»−3 2−6 4

–, B =

»−1 23 −2

–, C =

»1 46 1

»−3 2−6 4

– »−1 23 −2

»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)

»9 −10

18 −20

»−3 2−6 4

– »1 46 1

»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1

»9 −1018 −20

–Note:

Example 7: (also see the last example in the additional lecture notes)

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

–Note:

AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.

There will be no zero factor property for the multiplication of any two randommatrices.

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

–Note:

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

–Note:

AB = 0 despite the face that A 6= 0, B 6= 0.

In this case, the zero factor propertydose NOT hold.

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

–Note:

Consider

»1 22 4

–, B =

»−16 2

8 −1

»1 22 4

– »−16 2

8 −1

»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)

»0 00 0

–Note:

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

Solutions:

A is 3× 1, and B is 1× 3. Then AB should be 3× 3.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

Solutions: A is 3× 1, and B is 1× 3.

Then AB should be 3× 3.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

Solutions: A is 3× 1, and B is 1× 3. Then AB should be 3× 3.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Example 8:

Consider A =

24 1−12

35 , B =ˆ

1 3 2˜, then AB =?.

24 1−12

35 ˆ 1 3 2˜

24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2

24 1 3 2−1 −3 −22 6 4

Thank you!I will greatly appreciate it if you can tell me the typo that you find.

Calculus for Life Sciences · 2013-01-22 · Calculus for Life Sciences MAT 1332 C Winter 2010 Jing...

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