Chapter 2 Matrix Algebra
2.1 Addition, Scalar Multiplication, and Multiplication of Matrices
Addition of Matrices
Example 1
Scalar Multiplication of matrices
Negation and Subtraction
Multiplication of Matrices
Example 4
Example 5
Size of a Product Matrix
Theorem 2.1
Homework
2.2 Algebraic Properties of Matrix Operations
Arithmetic Operations
Example 10
Powers of Matrices
Example 12
Systems of Linear Equations
Idempotent and Nilpotent Matrices
Homework
2.3 Symmetric Matrices
Theorem 2.4: Properties of Transpose
Symmetric Matrix
Remark: If and only if
Example 17
Example 18
Homework
2.4 The Inverse of a Matrix
Theorem 2.5
Gauss-Jordan Elimination for finding the Inverse of a Matrix
Example 20
Example 21
Properties of Matrix Inverse
Theorem 2.6
Example 22
Elementary Matrices
Elementary Matrices
Notes for elementary matrices
Homework
1.01M
Category: mathematicsmathematics

Linear Algebra. Chapter 1. Matrix Algebra

1. Chapter 2 Matrix Algebra

Linear Algebra
Chapter 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

Definition
Let 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

Definition
Let 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

Definition
We 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

Definition
Let 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 1
1 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

1
2
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 an
m 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 Matrices
Definition
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 be
an 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 the
practical 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 -1
Let 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 -2
Let 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 scalar
multiplications 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

4
3 , 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.

Caution
In 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

Definition
If 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 2
4
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 follows
a11 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 teachers
of the practical classes.
Ch2_24

25. 2.3 Symmetric Matrices

Definition
The 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 sizes
of 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

Definition
A 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 that
the 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 of
the practical classes.

32. 2.4 The Inverse of a Matrix

Definition
Let 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 2
Determine 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, Then
1. ( 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 22
If 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 1
Solve 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

Definition
An 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 practical
classes.
Exercise
d b
1
a b
1
.
If A
, show that A
(ad bc) c a
c d
Ch2_44
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