Linear Algebr chapter3hts Determinants

Definition The determinant of a 2  2 matrix A is denoted |A| and is given by Observe that the determinant of a 2  2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A. Definition The determinant of a 2  2 matrix A is denoted |A| and is given by Observe that the determinant of a 2  2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A.

Chapter 3

Determinants

Linear Algebra

Observe that the determinant of a 2 × 2 matrix is given by the

different of the products of the two diagonals of the matrix.

The notation det(A) is also used for the determinant of A.

21 12 22

11 22

21

12

a a

2))

3(4()12(1

3 4

2)

det(A = − = × − × − = + =

Definition

Let A be a square matrix

The minor of the element a ij is denoted M ij and is the determinant

of the matrix that remains after deleting row i and column j of A.

The cofactor of a ij is denoted C ij and is given by

C ij = (–1)i+j M ij

Note that C ij = M ij or −M ij

10)

43()21(2

41 31

2

41 0 3

:of

−−

=

M a

3))

2(2()11(1

21 21

2

41 0 3

:of

−−

=

M a

Example 2

Solution

the following matrix A.

A

3)

3()1()

1(

:of

Cofactor a11 C11 = − 1+1M11 = − 2 =

10)

10(

)1()

1(

:of

Cofactor a32 C32 = − 3+2 M32 = − 5 − =

Definition

The determinant of a square matrix is the sum of the products

of the elements of the first row and their cofactors

These equations are called cofactor expansions of |A|.

n

n C a C

a C

a C

a A

n n A

C a C

a C

a C

a A

A

C a C

a C

a A

A

1 1 13

13 12

12 11

11

14 14 13

13 12

12 11

11

13 13 12

12 11

11

,is

If

4,4

isIf

3,3

isIf

++

++

=

×

++

+

=

×

++

=

×





1)(

1(1

43 1)

1(21

2 1

0)1(

13 13 12

12 11

11

−

−+

−+

−

=

++

= a C a C a C

A

6

622

)]

40()23[(

)]

41()13[(

2)]

21()10[(

−

= − + −

=

Theorem 3.1

The determinant of a square matrix is the sum of the products of

the elements of any row or column and their cofactors

ith row expansion:

jth column expansion: j j j j nj nj

in in i

i i

i

C a C

a C

a A

C a C

a C

a

A

++

+

=

++

1

2 2 1

12

1

A

Solution

66

012

)]

42()21[(

1)]

41()11[(

0)]

21()12[(

3

2

41 2

11

41 1

01

23

23 23 22

22 21

21

−

=+

−

−

=

++

= a C a C a C

A

40

12

Solution

6)

23(63

1)2(3

31

23

0)(

3)(

0)(

43 43 33

33 23

23 13

13

C C

C C

C a C

a C

a C

a

A

++

+

=

++

+

=

Example 6

Solve the following equation for the variable x.

72

1)(

1(

)2(x − − x + − =

x

Proceed to simplify this equation and solve for x.

3

or 2

0)

3)(

2(

06

71

2

2 2

−

=

=

−+ − − =

=++

−

x

x x

x x

x x x

There are two solutions to this equation, x = – 2 or 3

12 11

a a

31

23 22

21

13 12

11

a a

a

a a

a

a a

a

A

32 31

22 21

12 11

33 32

31

23 22

21

13 12

11

a a

a a

a a

a a

a

a a

a

a a

from products

(diagonal

right) left to

from products

(diagonal

A

13 22 31 11 23 32 12 21 33

32 21 13 31

23 12 33

22 11

a a a a

a a a

a a

a a a a

a a a

a a

−

−

−

+ +

=

⇒

Homework

Exercise 3.1 pages161-162:

3, 6, 9, 11, 13, 14

Let A be an n × n matrix and c be a nonzero scalar.

(a) If then |B| = c|A|.

Example 1

3 9

2

3 6

1

2 4

3

1

2

3 ) 3

( 3

3 2

3 0

1

2 0

3

3 9

2

3 6

1

2 4

3

C3 2

5 2

0

3 4

1 )

c

( 5 2

0

10 4

2

3 4

1 (b)

10 12

2

5 6

0

3 12

1 )

A

R

R +≈

Theorem 3.3

Let A be a square matrix A is singular if

(a) all the elements of a row (column) are zero

(b) two rows (columns) are equal

(c) two rows (columns) are proportional (i.e., Ri=cRj)

Proof

(a) Let all elements of the kth row of A be zero.

00

0

2 2 1

4 2 1

3 1

2 (b)

9 0

4

1 0

3

7 0

2 )

Solution

(a) All the elements in column 2 of A are zero Thus |A| = 0.

(b) Row 2 and row 3 are proportional Thus |B| = 0.

Theorem 3.4

Let A and B be n × n matrices and c be a nonzero scalar.

(a) |cA| = c n |A|.

cA

cRn cR

, , 2 , 1

(d)

A

A I

A A A

A ⋅ −1 = ⋅ −1 = =1 ⇒ −1 = 1

Example 4

the following determinants

(a) |3A| (b) |A2| (c) |5A t A–1|, assuming A–1 exists

Solution

(a) |3A| = (32)|A| = 9 × 4 = 36

(b) |A2| = |AA| =|A| |A|= 4 × 4 = 16

(c) |5A t A–1| = (52)|A t A–1| = 25|A t ||A–1| = 25 1 = 25

A A

A A A

A A A A

A A A

|A| = 0 ⇒ |AB| = |A||B| = 0

Thus the matrix AB is singular.

(⇐)

|AB| = 0 ⇒ |A||B| = 0 ⇒ |A| = 0 or |B| = 0

Thus AB being singular implies that either A or B is

singular

The inverse is not true

r q

p

f d

b

w v

u

r q p

e c

a

w v

u

r q

p

f e d

c b

a

+

=

+ +

+

Solution

v u

q

p f

e w

u

r

p d

c w

v

r

q b

a w

v u

r q

p

f e d c b

a

) (

) (

)

=

+ +

+

3.3 Numerical Evaluation of a

Determinant

Definition

A square matrix is called an upper triangular matrix if all the

elements below the main diagonal are zero

It is called a lower triangular matrix if all the elements above

the main diagonal are zero

triangular upper

1 0 0 0

9 0 0 0

5 3 2 0

7 0 4 1

, 9 0 0

5 1 0

2 8 3

1 8 5 4

0 2 0 7

0 0 4 1

0 0 0 8

, 8 9 3

0 1 2

0 0 7

find

, 5 0

0

4 3

0

9 1

nn

n n

nn

n

n

a a

a a

a a

a a

a a a a

a a

a a

a a a

a a

a a

4 44

3 34

33

22 11

3 33

2 23

22

11

2 22

1 12

11

0 0

0

0 0

0

0 0

5 ( 3 2

ol. A = × × − = −

S

Numerical Evaluation of a Determinant

8 1 0

5 1 0

1 4

2

R1 ) 2 ( 3 R

R1

R2 10

9 4

4 5

2

1 4

2

−

− + +

=

−

−

130

0

51

0

14

22

=

R

2613

)1(

2× − × = −

=

Solution (elementary row operations ）

Example 3

Evaluation the determinant

12

0 0

2 2

0 0

2 3

1 0

1 2

0 1

R1 R4

R1 ) 1 ( R3

R1 ) 2 ( R2

1 2 0 1

3 0 0 1

0 1 1 2

1 2 0 1

−

−

−

− +

− +

− +

=

−

−

6 0

0 0

2 2

0 0

2 3

1 0

1 2

0 1

=

126

)2()1(

1× − × − × =

=

Example 4

Evaluation the determinant

112

42

0

1 0

0

4 2

1

1 R ) 1 ( 3 R

R1

R2 11

2 2

5 2

1

4 2

1

−

−

− + +

0

3 2

0

4 2

1 ) 1 (

R3

1(21)1(− × × × − =

=

Example 5

Evaluation the determinant

15

6

20

11

00

03

00

52

00

20

11

R1)6(R4

R1)2(R3

R1

R2

15

66

43

22

32

11

20

11

−

−

−+

−++

3.4 Determinants, Matrix Inverse,

and Systems of Linear Equations

Definition

Let A be an n × n matrix and C ij be the cofactor of a ij

The matrix whose (i, j)th element is C ij is called the matrix of

cofactor of A

The transpose of this matrix is called the adjoint of A and is

denoted adj(A)

cofactor of

matrix

2 1

2 22

21

1 12

n

n n

C C

C

C C

C

C C

2 1

2 22

21

1 12

C C

C

C C

C

C C

C

nn n

n

n n

12

6 7

9

1 3

Theorem 3.6

Let A be a square matrix with |A| ≠ 0 A is invertible with

)(adj

i j

i

C a C

a C

a

C

C

C a

a a

A j

A i

j i

++

1

2

1

2 1

))adj(

of(column

)of(row

element

th ),

(

j i

A j

i

if 0

if

element

th )

Proof of Theorem 3.6

.2

2 1

(⇐) Theorem 3.6 tells us that if |A| ≠ 0, then A is invertible.

A–1 exists if and only if |A| ≠ 0.

1

10

2

1

4

|B| = 0 B is singular The inverse does not exist.

|C| = 0 C is singular The inverse does not exist.

|D| = 2 ≠ 0 D is invertible.

6 25

1 25

7 25

12 25

9 25

14

8 6

1

1 7

3

12 9

14 25

1 )

adj(

1

A A

Exercise 3.3 page 178-179: 4, 7.

Exercise

Show that if A = A-1, then |A| = ± 1

Show that if At = A-1, then |A| = ± 1.

Theorem 3.8

Let AX = B be a system of n linear equations in n variables

(1) If |A| ≠ 0, there is a unique solution

(2) If |A| = 0, there may be many or no solutions.

⇒ since A ≈…≈ C implies that if |A|≠0 then |C|≠0 (Thm 3.2)

⇒ the reduced echelon form of A is not I n

⇒ The solution to the system AX = B is not unique

⇒ many or no solutions

53

4

2 2

33

3 2

1

3 2

1

3 2

1

=+

+

−

=+

+

=

−+

x x

x

x x

x

x x

x

Solution

Since

01

Theorem 3.9 Cramer’s Rule

Let AX = B be a system of n linear equations in n variables such

that |A| ≠ 0 The system has a unique solution given by

Where A i is the matrix obtained by replacing column i of A with

B.

A

A x

A

A x

Proof

|A| ≠ 0 ⇒ the solution to AX = B is unique and is given by

B

A A

B A X

)(adj1

1

=

= −

x i , the ith element of X, is given by

)(

11

)]

(adjof

row[

1

2 2 1

1

2

1 2

1

ni n i

i

n

ni i

i i

C b C

b C

b A

b

b

b C

C

C A

B A

i A

x

++

Proof of Cramer’s Rule

the cofactor expansion of |A i|

in terms of the ith column

Example 5

Solving the following system of equations using Cramer’s rule

6 3

2

5

52

2

3

3 2

1

3 2

1

3 2

1

=+

+

−

=+

+

−

=+

+

x x

x

x x

x

x x

32

1 5 1

21 3 1 B A

It is found that |A| = –3 ≠ 0 Thus Cramer’s rule be applied We

21 3 2

3

6

21 2 1

3

A

Giving

Cramer’s rule now gives

9 ,

6 ,

3 2 3

1 = − A = A = −

A

33

9

,

23

6

,

13

3

2 2

A

A x

A

A x

The unique solution is x1 =1 ,x2 = −2 ,x3 = 3.

Example 6

equations has nontrivial solutions.Find the solutions for each

value of λ

0)

1(

2

0)

4(

)2(

2 1

2 1

=+

+

=+

+

+

x x

x

x

λ

λλ

Solution

homogeneous system

⇒ x1 = 0, x2 = 0 is the trivial solution

⇒ nontrivial solutions exist ⇒ many solutions

⇒

⇒ ⇒ ⇒

⇒ λ = – 3 or λ = 2

01

0 )

3 )(

2 ( λ − λ + =

0 6

2 + λ − =

λ

0 ) 4 (

2 ) 1 )(

2 ( λ + λ + − λ + =

λ = – 3 results in the system

This system has many solutions, x1 = r, x2 = r.

02

2

0

2 1

2 1

=

−

=+

−

x x

x x

λ = 2 results in the system

This system has many solutions, x2 3 1 = – 3r/2, x0 2 = r.

06

4

2 1

2 1

=+

=

+

x x

x x

Homework

Exercise 3.3 pages 179-180:

8, 12, 14, 15, 17.