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```Linear Algebra
Chapter 2
Matrices
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
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
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, there are
not of the same size, A + C does not exist.
Ch2_4
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
Let A   1  2 4.
7  2 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
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
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
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
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
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
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.
0mn
0
 0

0
0
0

0




zero matrix
0
0

0
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
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
 2
BO2  
 3
 2
BI 2  
 3
1  3 0 0 0 2 1  3


A
5
8 0 0 0 4 5
8
1 0
3 0
1 1

4 0
0  0


0  0
0  2

1  3
0
 O2

0
1
B
4
Ch2_12
Homework
Exercises: 5, 8, 11, 17 from page 71 to 73.
Exercise 17 p.72
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
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
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
5. (a + b)C = aC + bC
6. (ab)C = a(bC)
Ch2_14
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 zero matrix)
5. c(AB) = (cA)B = A(cB)
Note: AB BA in general.
Multiplication of matrices is not
commutative.
Ch2_15
Proof of Thm 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
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
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
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
, and C  
.



 2 4
 2 1
 3  2
 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
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
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  2B)  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
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
Idempotent and Nilpotent Matrices
(Exercises 24-30 p.83 and 14 p.93)
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
Homework
Exercises 3, 6, 12, 18, 27, 32, p.82 to
p.83
Ch2_24
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
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
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
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
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
Example 18
Let A be a symmetric matrix. Prove that A2 is symmetric.
Proof
( A2 )t  ( AA) t ( At At )  AA  A2
Ch2_30
Exercises 1, 2, 6, 7, 14, p.93 to p.94
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
 2
AB   1 2  3  1    1 0  I 2
3 4  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
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
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
Example 20
 1  1  2
Determine the inverse of the matrix A   2  3  5
3
5
 1
Solution
 1 1  2 1 0
[ A : I3 ]   2  3  5 0 1
3
5 0 0
 1
  1 1  2 1 0
(1)R2 0
1
1 2 1
3 1 0
0 2

1 0 0
0
 1 1  2
0 R2  (2) R10  1  1  2 1 0
3
1 0 1
1 R3  R1 0 2
0

3  1 0
 1 0 1
0 R1  R2 0 1 1
2  1 0
1 R3  (2)R2 0 0
1  3 2 1

0
1 1
1 0 0
R1  R3 0 1 0
5  3  1
R2  (1)R3 0 0 1  3
2 1
1 1
 0
Thus, A1   5  3  1.
 3
2
1
Ch2_35
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 1 5 1 0 0
1
2  1 1 0
[ A : I 3 ]   1 2 7 0 1 0 R2  (1)R1 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
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
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
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
Example 22
x1  x2  2 x3  1
Solve the system of equations 2 x1  3x2  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
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
Elementary Matrices
。 Elementary row operation
。 Elementary matrix
a b
R2  R3  g h

d e
a b
A  d e
 g h
c
f 
i 
5R2
a
5d

 g
c  1 0 0
i   0 0 1  A  E1 A
f  0 1 0
b c  1 0 0
5e 5 f   0 5 0  A  E2 A
h
i  0 0 1
b
 a
d  2a e  2b
R2+ 2R1 
 g
h
c  1 0 0
f  2c   2 1 0  A  E3 A
i  0 0 1
Ch2_42
Notes for elementary matrices
Each elementary matrix is invertible.
Example 24
I  E1  E1  I , i.e., E2 E1  I
R1 2 R 2
1 0 0
I  0 1 0
0 0 1
R1 2 R 2
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
Homework
Exercises 6, 9, 14, 15, 16, 19, 21, 32 from p.105 to 107.
Exercise 7
 d  b
1
a b 
1
.
If A  
, show that A 



(ad  bc)  c a 
c d 
Ch2_44
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