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# Linear Algebra. Chapter 1. Matrix Algebra

## 1. Chapter 2 Matrix Algebra

Linear AlgebraChapter 2

Matrix Algebra

## 2. 2.1 Addition, Scalar Multiplication, and Multiplication of Matrices

• aij: the element of matrix A in row i and column j.• For a square n n matrix A, the main diagonal is:

a11 a12

a

a22

21

A

an1 an 2

a1n

a2 n

ann

Definition

Two matrices are equal if they are of the same size and if their

corresponding elements are equal.

Thus A = B if aij = bij i, j.

( for every, for all)

Ch2_2

## 3. Addition of Matrices

DefinitionLet A and B be matrices of the same size.

Their sum A + B is the matrix obtained by adding together the

corresponding elements of A and B.

The matrix A + B will be of the same size as A and B.

If A and B are not of the same size, they cannot be added, and we

say that the sum does not exist.

Thus if C A B, then cij aij bij i,j .

Ch2_3

## 4. Example 1

4 7 , B 2 5 6 , and C 5 4 .Let A 1

8

0 2 3

3 1

2 7

Determine A + B and A + C, if the sum exist.

Solution

4 7 2 5 6

(1) A B 1

8

0 2 3 3 1

4 5 7 6

1 2

0 3 2 1 3 8

3 9 1 .

3 1 11

(2) Because A is 2 3 matrix and C is a 2 2 matrix, they are

not of the same size, A + C does not exist.

Ch2_4

## 5. Scalar Multiplication of matrices

DefinitionLet A be a matrix and c be a scalar. The scalar multiple of A by c,

denoted cA, is the matrix obtained by multiplying every element

of A by c. The matrix cA will be the same size as A.

Thus if B cA, then bij caij i, j.

Example 2

1 2 4

Let A

.

7 3 0

3 1 3 ( 2) 3 4 3 6 12

3A

.

3 7 3 ( 3) 3 0 21 9 0

Observe that A and 3A are both 2 3 matrices.

Ch2_5

## 6. Negation and Subtraction

DefinitionWe now define subtraction of matrices in such a way that makes it

compatible with addition, scalar multiplication, and negative. Let

A – B = A + (–1)B

Example 3

Suppose A 5 0 2 and B 2 8 1 .

3 6 5

0 4 6

5 2 0 8 2 ( 1) 3 8 1

A B

.

2 11

3 0 6 4

5 6 3

Ch2_6

## 7. Multiplication of Matrices

DefinitionLet the number of columns in a matrix A be the same as the

number of rows in a matrix B. The product AB then exists.

Let A: m n matrix, B: n k matrix,

The product matrix C=AB has elements

cij ai1

ai 2

b1 j

b

2j

ain

ai1b1 j ai 2b2 j ainbnj

bnj

C is a m k matrix.

If the number of columns in A does not equal the number of row B,

we say that the product does not exist.

Ch2_7

## 8. Example 4

0 11 3

5

Let A

,B

, and C 6 2 5 .

2 0

3 2 6

Determine AB, BA, and AC, if the products exist.

Solution.

AB 1

2

5

1 3

3

5

2

0

3

3 5

0

3

0

2

0

1 3

2

0

2 0

2

1

6

1

1 3

6

1

2 0

6

(1 5) (3 3) (1 0) (3 ( 2)) (1 1) (3 6)

(2 5) (0 3) (2 0) (0 ( 2)) (2 1) (0 6)

14 6 19

.

10

0

2

BA and AC do not exist.

Note. In general, AB BA.

Ch2_8

## 9. Example 5

12

Let A 7

0 and B 1 0 . Determine AB.

3 5

3 2

2 1 1

2 1 0

3

5

1

2

1

0

1

0

7 0 5

AB 7

0

7 0

3

5

3

3 2

1

0

3 2

3 2

3

5

5

2 3 0 5 1

7 0 0 0 7

0

3 6 0 10 3 10

Example 6

Let C = AB, A 2 1 and B 7 3 2 Determine c23.

3 4

5 0 1

2 ( 3 2) (4 1) 2

3

4

c23

1

Ch2_9

## 10. Size of a Product Matrix

If A is an m r matrix and B is an r n matrix, then AB will be anm n matrix.

A

B

= AB

m r

r n

m n

Example 7

If A is a 5 6 matrix and B is an 6 7 matrix.

Because A has six columns and B has six rows. Thus AB exits.

And AB will be a 5 7 matrix.

Ch2_10

## 11.

Special MatricesDefinition

A zero matrix is a matrix in which all the elements are zeros.

A diagonal matrix is a square matrix in which all the elements

not on the main diagonal are zeros.

An identity matrix is a diagonal matrix in which every diagonal

element is 1.

0 0 0

0mn 0 0 0

0 0 0

zero matrix

a11 0

0 a

22

A

0

0

0

0

ann

1 0 0

I n 0 1 0

0 0 1

identity matrix

diaginal matrix A

Ch2_11

## 12. Theorem 2.1

Let A be m n matrix and Omn be the zero m n matrix. Let B bean n n square matrix. On and In be the zero and identity n n

matrices. Then

A + Omn = Omn + A = A

BOn = OnB = On

BIn = InB = B

Example 8

Let A 2 1 3 and B 2 1 .

8

4 5

3 4

2

A O23

4

1 3

0 0

1 3

5

0

5

0

8 0

2

0 4

2 1 0 0 0 0

BO2

O2

3 3 0 0 0 0

2 1 1 0 2 1

B

BI 2

3 4 0 1 3 4

A

8

Ch2_12

## 13. Homework

Exercises will be given by the teachers of thepractical classes.

Exercise

Let A be a matrix whose third row is all zeros. Let B be any

matrix such that the product AB exists.

Prove that the third row of AB is all zeros.

Solution

b1i

b1i

b

b

( AB) 3i [a31 a32 a3n ] 2i [0 0 0] 2i 0, i.

bni

bni

Ch2_13

## 14. 2.2 Algebraic Properties of Matrix Operations

Theorem 2.2 -1Let A, B, and C be matrices and a, b, and c be scalars. Assume that the

size of the matrices are such that the operations can be performed.

Properties of Matrix Addition and scalar Multiplication

1. A + B = B + A

Commutative property of addition

2. A + (B + C) = (A + B) + C Associative property of addition

3. A + O = O + A = A

(where O is the appropriate zero matrix)

4. c(A + B) = cA + cB

Distributive property of addition

5. (a + b)C = aC + bC

Distributive property of addition

6. (ab)C = a(bC)

Ch2_14

## 15.

Theorem 2.2 -2Let A, B, and C be matrices and a, b, and c be scalars. Assume that the

size of the matrices are such that the operations can be performed.

Properties of Matrix Multiplication

1. A(BC) = (AB)C

Associative property of multiplication

2. A(B + C) = AB + AC

Distributive property of multiplication

3. (A + B)C = AC + BC

Distributive property of multiplication

4. AIn = InA = A

(where In is the appropriate identity matrix)

5. c(AB) = (cA)B = A(cB)

Note: AB BA in general.

Multiplication of matrices is not

commutative.

Ch2_15

## 16.

Proof of Theorem 2.2 (A+B=B+A)Consider the (i,j)th elements of matrices A+B and B+A:

( A B)ij aij bij bij aij ( B A)ij .

A+B=B+A

Example 9

Let A 1 3 , B 3 7 , and C 0 2 .

1

4 5

8

5 1

A B C 1 3 3 7 0 2

1 5 1

4 5 8

1 3 0 3 7 2 4 6 .

5

4 8 5 5 1 1 9

Ch2_16

## 17. Arithmetic Operations

If A is an m r matrix and B is r n matrix, the number of scalarmultiplications involved in computing the product AB is mrn.

Consider three matrices A, B and C such that the product

ABC exists.

Compare the number of multiplications involved in the

two ways (AB)C and A(BC) of computing the product ABC

Ch2_17

## 18. Example 10

43 , and C 1 .

Let A 1 2 , B 0 1

Compute ABC.

3 1

1 0 2

0

Solution.

Which method is better?

Count the number of multiplications.

(1) (AB)C

3 2 1 1 .

AB 1 2 0 1

3 1 1 0 2 1 3 11

2 6+3 2

4 9

=12+6=18

2

1

1

1 .

( AB)C

1 3 11 0 1

(2) A(BC)

4

0

1

3

1 1

BC

1 0 2 0 4

A( BC ) 1 2 1 9 .

3 1 4 1

3 2+2 2

=6+4=10

A(BC) is better.

Ch2_18

## 19.

CautionIn algebra we know that the following cancellation laws apply.

If ab = ac and a 0 then b = c.

If pq = 0 then p = 0 or q = 0.

However the corresponding results are not true for matrices.

AB = AC does not imply that B = C.

PQ = O does not imply that P = O or Q = O.

Example 11

1 2

1 2

3 8

(1) Consider t he matrices A

, B 2 1 , and C 3 2 .

2

4

3 4

Observe that AB AC

, but B C.

6 8

1 2

2 6

(2) Consider t he matrices P

, and Q

.

4

2

1 3

Observe that PQ O, but P O and Q O.

Ch2_19

## 20. Powers of Matrices

DefinitionIf A is a square matrix, then

Ak

AA

A

k times

Theorem 2.3

If A is an n n square matrix and r and s are nonnegative

integers, then

1. ArAs = Ar+s.

2. (Ar)s = Ars.

3. A0 = In (by definition)

Ch2_20

## 21. Example 12

1 24

If A

,

compute

A

.

0

1

Solution

1 2 1 2 3 2

A

1

0

1

0

1

2

2

3 2 3 2 11 10

A

.

2 1

2 5

6

1

4

Example 13 Simplify the following matrix expression.

A( A 2 B ) 3B ( 2 A B ) A2 7 B 2 5 AB

Solution

A( A 2 B) 3B (2 A B ) A2 7 B 2 5 AB

A2 2 AB 6 BA 3B 2 A2 7 B 2 5 AB

3 AB 6 BA 4 B 2

We can’t add the two matrices

Ch2_21

## 22. Systems of Linear Equations

A system of m linear equations in n variables as followsa11 x1 a1n xn b1

am1 x1 amn xn bm

Let

a11 a1n

x1

b1

A , X , and B

am1 amn

xn

bm

We can write the system of equations in the matrix form

AX = B

Ch2_22

## 23. Idempotent and Nilpotent Matrices

Definition(1) A square matrix A is said to be idempotent if A2=A.

(2) A square matrix A is said to nilpotent if there is a

p

positive integer p such that A =0. The least integer p such that

Ap=0 is called the degree of nilpotency of the matrix.

Example 14

3 6 2 3 6

(1) A

,A

A.

1 2

1 2

3 9 2 0 0

(2) B

,B

. The degree of nilpotency : 2

1 3

0 0

Ch2_23

## 24. Homework

Exercises will be given by the teachersof the practical classes.

Ch2_24

## 25. 2.3 Symmetric Matrices

DefinitionThe transpose of a matrix A, denoted At, is the matrix whose

columns are the rows of the given matrix A.

i.e., A : m n At : n m, ( At )ij Aji i, j.

Example 15

A 2 7 , B 1 2 7 , and C 1 3 4 .

6

8 0

4 5

1

1

4

t

t

C 3 .

At 2 8

B

2

5

0

7

4

7 6

Ch2_25

## 26. Theorem 2.4: Properties of Transpose

Let A and B be matrices and c be a scalar. Assume that the sizesof the matrices are such that the operations can be performed.

1. (A + B)t = At + Bt

Transpose of a sum

2. (cA)t = cAt

Transpose of a scalar multiple

3. (AB)t = BtAt

Transpose of a product

4. (At)t = A

Ch2_26

## 27. Symmetric Matrix

DefinitionA symmetric matrix is a matrix that is equal to its transpose.

A At , i.e., aij a ji i, j

Example 16

5

2

5 4

match

1

0 1 4 0

8

1 7

2

4 8 3 4

0 2

4

7

3

9

3

2 3

9 3

6

match

Ch2_27

## 28. Remark: If and only if

Let p and q be statements.Suppose that p implies q (if p then q), written p q,

and that also q p, we say that

“p if and only if q” (in short iff )

Ch2_28

## 29. Example 17

Let A and B be symmetric matrices of the same size. Prove thatthe product AB is symmetric if and only if AB = BA.

Proof

*We have to show (a) AB is symmetric AB = BA,

and the converse, (b) AB is symmetric AB = BA.

( ) Let AB be symmetric, then

AB= (AB)t

by definition of symmetric matrix

= BtAt

by Thm 2.4 (3)

= BA

since A and B are symmetric

( ) Let AB = BA, then

(AB)t = (BA)t

= AtBt

by Thm 2.4 (3)

= AB

since A and B are symmetric

Ch2_29

## 30. Example 18

Let A be a symmetric matrix. Prove that A2 is symmetric.Proof

( A ) ( AA) ( A A ) AA A

2 t

t

t

t

2

Ch2_30

## 31. Homework

Exercises will be given by the teachers ofthe practical classes.

## 32. 2.4 The Inverse of a Matrix

DefinitionLet A be an n n matrix. If a matrix B can be found such that

AB = BA = In, then A is said to be invertible and B is called the

inverse of A. If such a matrix B does not exist, then A has no

inverse. (denote B = A 1, and A k=(A 1)k )

Example 19

2 1

Prove that the matrix A 1 2 has inverse B 3 1 .

3 4

2

2

Proof

1 1 0

2

1

2

1

AB

I

3 4 32 2 0 1 2

1 1 2 1 0

2

BA 3 1

I2

3

4

0

1

2

2

Thus AB = BA = I2, proving that the matrix A has inverse B.

Ch2_32

## 33. Theorem 2.5

The inverse of an invertible matrix is unique.Proof

Let B and C be inverses of A.

Thus AB = BA = In, and AC = CA = In.

Multiply both sides of the equation AB = In by C.

C(AB) = CIn

Thm2.2

(CA)B = C

InB = C

B=C

Thus an invertible matrix has only one inverse.

Ch2_33

## 34. Gauss-Jordan Elimination for finding the Inverse of a Matrix

Let A be an n n matrix.1. Adjoin the identity n n matrix In to A to form the matrix

[A : In].

2. Compute the reduced echelon form of [A : In].

If the reduced echelon form is of the type [In : B], then B is

the inverse of A.

If the reduced echelon form is not of the type [In : B], in that

the first n n submatrix is not In, then A has no inverse.

An n n matrix A is invertible if and only if its reduced echelon

form is In.

Ch2_34

## 35. Example 20

1 1 2Determine the inverse of the matrix A 2 3 5

3

5

1

Solution

1 1 2 1 0 0

1

[ A : I 3 ] 2 3 5 0 1 0 R2 ( 2) R1 0

3

5 0 0 1 R3 R1 0

1

1 1 2 1 0 0

1

( 1)R2 0

1

1 2 1 0 R1 R2 0

3 1 0 1 R3 ( 2)R2 0

0 2

1 0 0

0

1

1

R1 R3 0 1 0 5 3 1

R2 ( 1)R3 0 0 1 3 2 1

1 1

0

Thus, A 5 3 1 .

2

1

3

1 2

1 0 0

1 1 2 1 0

2

3

1 0 1

0 1

3 1 0

1 1

2 1 0

0

1 3 2 1

1

Ch2_35

## 36. Example 21

Determine the inverse of the following matrix, if it exist.1 1 5

A 1 2 7

2 1 4

Solution

1

5

1 0 0

1 1 5 1 0 0

1

[ A : I 3 ] 1 2 7 0 1 0 R2 ( 1)R1 0

1

2 1 1 0

2 1 4 0 0 1 R3 ( 2)R1 0 3 6 2 0 1

2 1 0

1 0 3

R1 ( 1)R2 0 1 2 1 1 0

R3 3R2 0 0 0 5 3 1

There is no need to proceed further.

The reduced echelon form cannot have a one in the (3, 3) location.

The reduced echelon form cannot be of the form [In : B].

Thus A–1 does not exist.

Ch2_36

## 37. Properties of Matrix Inverse

Let A and B be invertible matrices and c a nonzero scalar, Then1. ( A 1 ) 1 A

1 1

1

2. (cA) A

c 1 1

1

3. ( AB) B A

4. ( An ) 1 ( A 1 ) n

5. ( At ) 1 ( A 1 ) t

Proof

1. By definition, AA 1=A 1A=I.

2. (cA)( 1c A 1 ) I ( 1c A 1 )(cA)

3. ( AB)( B 1 A 1 ) A( BB 1 ) A 1 AA 1 I ( B 1 A 1 )( AB)

1

1

1 n n

4. An ( A 1 ) n

A

A

A

A

I

(

A

) A

n times

n times

5. AA 1 I , ( AA 1 ) t ( A 1 ) t At I ,

A 1 A I , ( A 1 A) t At ( A 1 ) t I ,

Ch2_37

## 38.

Example 22If A 4 1 , then it can be shown that A 1 1 1 . Use this

3 1

3 4

informatio n to compute ( At ) 1.

Solution

t

1 1

1 3

(A ) (A )

.

3 4 1 4

t 1

1 t

Ch2_38

## 39. Theorem 2.6

Let AX = B be a system of n linear equations in n variables.If A–1 exists, the solution is unique and is given by X = A–1B.

Proof

(X = A–1B is a solution.)

Substitute X = A–1B into the matrix equation.

AX = A(A–1B) = (AA–1)B = In B = B.

(The solution is unique.)

Let Y be any solution, thus AY = B. Multiplying both sides of this

equation by A–1 gives

A–1A Y= A–1B

In Y= A–1B

Y = A–1B. Then Y=X .

Ch2_39

## 40. Example 22

x1 x2 2 x3 1Solve the system of equations 2 x1 3 x2 5 x3 3

x1 3 x2 5 x3 2

Solution

This system can be written in the following matrix form:

1 1 2 x1 1

2 3 5 x2 3

3

5 x3 2

1

If the matrix of coefficients is invertible, the unique solution is

1

x1 1 1 2 1

x2 2 3 5 3

x 1

3

5 2

3

This inverse has already been found in Example 20. We get

x1 0

1 1 1 1

x2 5 3 1 3 2

x 3

2

1 2 1

3

The unique solution is x1 1, x2 2, x3 1.

Ch2_40

## 41. Elementary Matrices

DefinitionAn elementary matrix is one that can be obtained from the

identity matrix In through a single elementary row operation.

Example 23

1 0 0

I 3 0 1 0

0 0 1

R2 R3

5R2

R2+ 2R1

1 0 0

E1 0 0 1

0 1 0

1 0 0

E2 0 5 0

0 0 1

1 0 0

E3 2 1 0

0 0 1

Ch2_41

## 42. Elementary Matrices

。 Elementary row operation。 Elementary matrix

a

g

R2 R3

d

a

A d

g

b

e

h

c

f

i

5R2

a

5d

g

c 1 0 0

i 0 0 1 A E1 A

f 0 1 0

b

h

e

c 1 0 0

5e 5 f 0 5 0 A E2 A

h

i 0 0 1

b

b

a

d 2a e 2b

R2+ 2R1

g

h

1 0 0

f 2c 2 1 0 A E3 A

i 0 0 1

c

Ch2_42

## 43. Notes for elementary matrices

Each elementary matrix is invertible.Example 24

I E1 E1 I , i.e., E2 E1 I

R1 2 R 2

R1 2 R 2

1 0 0

I 0 1 0

0 0 1

1 2 0

E1 0 1 0

0 0 1

1 2 0

E2 0 1 0

0 0 1

If A and B are row equivalent matrices and A is

invertible, then B is invertible.

Proof

If A … B, then

B=En … E2 E1 A for some elementary matrices En, … , E2 and E1.

So B 1 = (En … E2 E1A) 1 =A 1E1 1 E2 1 … En 1.

Ch2_43

## 44. Homework

Exercises will be given by the teachers of the practicalclasses.

Exercise

d b

1

a b

1

.

If A

, show that A

(ad bc) c a

c d

Ch2_44