Properties of a Determinant

8/10/2019 Properties of a Determinant

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Chapter 2: Determinants 17

SECTION BProperties of a Determinant

By the end of this section you will be able to

prove that the determinant of a triangular or diagonal matrix is the product of theleading diagonal entries

evaluate determinants of a triangular and diagonal matrix prove properties of determinant of an elementary matrix establish certain properties of determinants of other matrices

You will need to remember the definition of a determinant of a matrix and the

technique to evaluate it by using cofactors. This section is a difficult section to follow

because we need to prove a number of propositions and some of these rely on the

results of chapter 1.

B1 Revision of Properties of a Determinant

In the last section we established certain properties of the determinant of a matrix suchas:

Proposition (2.10). If a square nby n matrix A consists of two identical rows then

.( )det 0=A

Further properties were also established in the associated Exercise 2(a) and we give

these reference numbers.

Proposition (2.14). Let Abe a square matrix then ( ) ( )det detT =A A . What does thismean?

The determinant of the transposed matrix is the same as the determinant of the initial

matrix.

Proposition (2.15). Let Bbe a matrix obtained from matrix Aby multiplying onerow(or column) of Aby a non-zero scalar thenk

( ) ( )det detk=B A We will restate this result later in this section.

Proposition (2.16). Let Abe a square nby nmatrix and kbe a scalar then

( ) ( )det detnk k=A A

What does this mean?

The determinant of the scalar multiplication kAis times the determinant of the

matrix A.

nk

Note that since

( ) ( )det det

T =A A , expanding along a row or column is equivalent.

Generally propositions about the determinant of the matrix with the word row can be

swapped by the word columnbecause ( ) ( )det detT =A A .B2 Determinant Properties of Particular Matrices

We can find determinants of particular matrices such as triangular matrices . What are

triangular matrices?

Definition (2.17). A triangular matrix is a nby n matrix where all entries to one side of

the leading diagonal are zero.

What is meant by the leading diagonal?

Leading diagonal of a matrix is the entries going from the top left hand corner to the

bottom right hand corner of the matrix,For example, the following are triangular matrices:


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(a)

1 2 3 1 0 0

0 4 5 and (b) 2 3 0

0 0 6 4 5 6

(a) is an example of an upper triangular matrix.

Leading

Diagonal

(b) is an example of a lower triangular matrix.Another type of matrix is a diagonal matrix.Do you know what is meant by a diagonal

matrix?

Definition (2.18). A diagonal matrix is a nby n matrix where allentries to both sides of

the leading diagonal are zero.

Can you think of an example of a diagonal matrix?

The identity matrix . Another example is .

1 0 0

0 1 0

0 0 1

=

I

1 0 0

0 2 0

0 0 3

A diagonal matrix is both an upper and lower triangular matrix.Triangular and diagonal matrices have the following property.

Proposition (2.19). The determinant of a triangular or diagonal matrix is a product of

the entries along the leading diagonal.

What does this proposition mean?

Let Abe an upper triangle matrix such that

11 12 1

22 20

0 0

n

n

nn

a a a

a a

a

=

A

then .( ) 11 22 33det nna a a a= A We prove it for the upper triangular matrix and the proof for the lower triangular matrix

is similar. Since the diagonal matrix is a particular upper (or lower) triangular matrixtherefore the proof of the diagonal matrix follows from the upper triangular result.

We use proof by induction. Remember the method of proof by induction is threefold.

1. Prove it for a base (or for some other base1 or 2n n= = 0n k= ).

2. Assume it is true for n k.=3. Prove it for 1n k= + .Proof.For a 2 by 2 matrix we have

11 12

11 22 11 22

22

det 00

a aa a a a

a

= =

Hence the determinant of a 2 by 2 upper triangular matrix is the product of entries in

the leading diagonal. Therefore the result is true for 2n= .Assume it is true for n , that isk=

11 12 1

22 2

11 22 33

0det

0 0

k

k

kk

kk

a a a

a aa a a a

a

=

()


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We need to prove the result for 1n k= + , that is a matrix of size 1k+ by . For1k+1n k= + we can find the determinant by expanding along the bottom row because all

entries in the bottom row are zero apart from ( )( )1 1k ka + + :

( )

( )

( )( )

( )( ) ( )

( )( )

( ) ( )( )

( )( )

11 22 33

11 12 1 111 12 1

22 1 12 1 22 2

1 1

1 1

by ()

2 2

11 22 331 1

11 22 33 1 1

0 0det 1 det

0 00 0

1

kk

kk

k kk k

k k

kkk k

a a a a

k

kkk k

kk k k

a a a a a a

a a a aa

aa

a a a a a

a a a a a

+

+ + ++

+ +

+ +

=

+

+ +

+ +

=

=

=

Remember ( ) ( ) ( )2 2 2 1

1 1k k+ +

= = 1because the index ( )2 1k+ is even.

Hence the determinant of an upper triangular matrix of size 1k+ by is the1k+product of the entries in the leading diagonal. We have proven the case for .1n k= +Therefore by induction we have shown that the determinant of an upper triangularmatrix is the product of the entries along the leading diagonal.

Example 10Find the determinants of the following matrices:

(a) (b) (c)

1 2 3 4

0 5 6 7

0 0 8 9

0 0 0 10

=

U

1 0 0 0

3 2 0 0

11 8 5 0

89 9 3 2

=

L

2 0 0 0 0

0 6 0 0 0

0 0 4 0 0

0 0 0 3 0

0 0 0 0 10

=

D

Solution

We use the result of the above Proposition (2.19) because all 3 matrices are uppertriangular, lower triangular and diagonal respectively.

In each case the determinant is the product of the leading diagonal entries.

(a) ( )det 1 5 8 10 400= =U

(b) ( ) ( )det 1 2 5 2 20= = L

(c) ( ) ( )det 2 6 4 3 10 1440= =

D You may like to check these answers by using MATLAB. The MATLAB command for

determinant is det().

B3 Determinant Properties of Elementary Matrices

Do you remember what an elementary matrix is?

An elementary matrixis a matrix obtained by a singlerow operation on the identity

matrix I. Examples of 3 by 3 elementary matrices are

1 0 0

0 5 0

0 0 1

, and

0 0 1

0 1 0

1 0 0

1 0 5

0 1 0

0 0 1

There are three different types of elementary matrices.


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1. An elementary matrix E obtained from the identity matrix, I, by multiplying a

row by a non-zero scalar .k

2. An elementary matrix E obtained from the identity matrix, I, by adding (or

subtracting) one row to another.3. An elementary matrix E obtained from the identity matrix, I, by interchanging

two rows (or columns).We can evaluate the determinant of these three different elementary matrices by using

the following proposition.

Proposition (2.20). Let Ebe an elementary matrix.(a) If the elementary matrix E is obtained from the identity matrix Iby multiplying a

row by a non-zero scalar kthen ( )det k=E .

(b)If the elementary matrix E is obtained from the identity matrix Iby adding (or

subtracting) a multiple of one row to another then ( )det 1=E .

(c) If the elementary matrix E is obtained from the identity matrix Iby interchanging

two rows (or columns) then ( )det 1= E .

Proof.(a) Remember the determinant of the identity is 1 therefore by the above Proposition

(2.15) with one row of the identity multiplied by non-zero k we have

( ) ( )det det 1k k k= = E I =

(b) Remember the identity matrix is a diagonal matrix. If we add a multiple of one row

of the identity matrix to another then we have a triangular matrix and by the aboveproposition (2.19) the determinant of a triangular matrix is the product of the elements

along the main diagonal, which is 1 1 1 1 1 = . Hence ( )det 1=E .

(c) We have to prove it for the case when two rows have been interchanged. This is a

more complex proof and is by induction.We first prove the result for :2n=

( ) 2Rows of have0 1

det det1 0 been interchanged

0 1 1

=

= =

IE

Hence if we interchange the two rows of the 2 by 2 identity matrix then the determinantis .1Assume the result is true for , that is by k elementary matrix with rows in k= k kE

and j interchanged, .( )det 1k = E

We need to prove the result for 1n k= + . Let1k+E be the 1 by 1k k+ + elementary

matrix with rows i and j of the identity matrix interchanged.

1

1 0 0

0 0 1 0

0 1 0

0 1 0 0

1

0 0 1

k+

=

E

(2.15) de where Bis obtained by multiplying one row of Aby k( ) ( )

jth row

ith row

t detk=B A


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To find the determinant of this matrix we can expand along the kth row where kth row

is not oneof ith or jth row. Note that in the kth row allthe entries should be zeros apart

from the diagonal element which is equal to 1. Therefore the determinant of thiskke

matrix is1k+E

( ) ( ) (1det 1 detk k

k

+

+ = E )kE (*)Why?

Because if you delete the elements containing the row and column containing the entry

kke then the remaining matrix is the by k elementary matrixwith rows i and jk

interchanged, which is . What is the determinant of ?kE kE

By our induction hypothesis we have ( )det 1k = E . Substituting this into (*) gives

( ) ( ) ( )

( ) ( ) ( )

1

2 2

det 1 1

1 1 1 Because 1 1

k k

k

k k

+

+ =

= = =

E

Hence we have proven that the determinant of an elementary matrix of size by1k+1k+ which has two rows interchanged is 1 .

Therefore by induction we have our result that if the elementary matrix E is obtained

from the identity matrix Iby swapping two rows (or columns) then .( )det 1= E

Note that we can use this Proposition (2.20) to find the determinants of elementary

matrices. Summarizing this Proposition (2.20) we have that the determinant of an

elementary matrix E is given by

(2.21) ( )

1 if a multiple of one row is added to another

det 1 if two rows have been interchanged

if a row has been multiplied by non-zerok k

=

E

Example 11

Find the determinants of the following elementary matrices:

(a) (b) (c)

1 0 0

0 5 0

0 0 1

=

A

0 0 1

0 1 0

1 0 0

=

B

1 0 5

0 1 0

0 0 1

=

C

Solution

Each of these is an elementary matrix so we can use the above Result (2.21).(a) Since this A is the elementary matrix obtained from the identity matrix by

multiplying the middle row by we have, by Result (2.21) with5 5k= ,

( )det 5= A

(b)How is matrixB obtained from the identity matrix?

By interchanging top and bottom rows, therefore by the middle line of (2.21) we have

( )det 1= B

(c)How is matrix C obtained from the identity matrix?

By subtracting 5 times the bottom row from the top row. Hence by the top line of

Result (2.21) we have

( )det 1=C


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B4 Determinant Properties of Other MatricesWe can extend Proposition (2.20) or result (2.21) to any square matrix Aas the

following proposition states.

Proposition (2.22). Let Bbe a matrix obtained from the matrix Aby

(a) multiplying a row (or column) by a non-zero scalar k. In this case

( ) ( )det detk=B A .(b) adding (or subtracting) a multiple of one row to another. In this case

( ) ( )det det=B A .

(c) interchanging two rows (or columns). In this case ( ) (det det= B A) .

Proof.

Part (a) was proven in Exercise 2(a) and is really Proposition (2.11) stated earlier in this

section. See Exercise 2(b) for proofs of parts (b) and (c).

We can summarize this into the following result.

(2.23) ( )

( )

( )

( )

det if a multiple of one row is added to another

det det if two rows have been interchanged

det if a row has been multiplied by non-zerok k

=

A

B A

A

In the next example we apply this result (2.23) to find the determinant of a matrix

which has fractional entries.

Example 12

Find the determinant of the following matrix:

1 21

23 23

1 1 5

2 6 61 3 1

11 11 11

=

A

Solution

How can we find the determinant of this matrix?

We can find the determinant of another matrix which is matrix A with top row

multiplied by 23, second row multiplied by 6 and bottom row multiplied by 11.

How is the determinant of this new matrix, call itB say, related to the determinant of

matrixA?

( ) ( ) ( )

1 2 23

23 6 11 det det det 3 1 5

1 3 1

= =

A B ()

We can find the determinant of matrix B as in the last section:

( ) ( )

( ) ( ) ( )

( )

1 2 231 5 3 5 3 1

det det 3 1 5 det 2 det 23det3 1 1 1 1 3

1 3 1

1 15 2 3 5 23 9 1

16 16 23 8 152

= = +

= + +

= + =

B

Substituting this, , into () gives( )det 152=B


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( ) ( )

( )( )

23 6 11 det 152

152 152 76det

23 6 11 1518 759

=

= =

A

A =

Hence ( )

76

det 759=

A .

For larger size matrices it is normally easier to convert these into triangular matrices by

applying row operations. Why carry out this conversion?

Because the determinant of a triangular matrix is just the product of the entries on the

leading diagonal. The following two examples shows how we use this approach.

Example 13

Find the determinant of the following matrix by using row operations:

1 2 2 4

7 8 3 0

3 2 0 0

1 0 0 0

=

A

Solution

Can we convert this into a triangular matrix?

Yes by using row operations. Note that the given matrix A is not a triangular matrix

because both sides of the leading diagonal contain non-zero entries. You will need to

recall your work from chapter 1 on row operations.How can we convert the matrixA

into a triangular matrix?

First we label the rows of matrix Aand then we apply row operations:

1

2

3

4

1 2 2 4

7 8 3 0

3 2 0 0

1 0 0 0

R

R

R

R

Interchanging rows1R and 4R , 2R and 3R we have

4

3

2

1

1 0 0 0

3 2 0 0

7 8 3 0

1 2 2 4

R

R

R

R

What is the determinant of this matrix?We have a lower triangular matrix so the determinant is the product of the entries on

the leading diagonal, that is 1 . What is the determinant of the given2 3 4 24 =matrixA?

The bottom matrix is obtained from matrix Aby interchanging rows1R and 4R , 2R

and3

R .How does interchanging rows affect the determinant of the matrix?

By (2.23) interchanging rows multiplies the determinant by 1 . Since we have twointerchanges therefore the determinant of the matrix Ais given by

( ) ( ) ( )1 1 det 2 =A 4 which gives ( )det 24=A

(2.23) ( ) ( )det det if two rows have been interchanged = B A


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Example 14

Find the determinant of the following matrix:

5 1 1 1 1

1 5 1 1 1

5 1 5 1 11 5 1 5 1

5 1 5 1 5

=

A

Solution

Labelling the rows of this matrix we have

1

2

3

4

5

5 1 1 1 1

1 5 1 1 1

5 1 5 1 1

1 5 1 5 1

5 1 5 1 5

R

R

R

R

R

Executing the following row operations:

1

2 2 1

3 3 1

4 4 2

5 5 3

5 1 1 1 1

* 4 4 0 0 0

* 0 0 4 0 0

* 0 0 0 4 0

* 0 0 0 0 4

R

R R R

R R R

R R R

R R R

= =

= =

Carrying out the row operation2

5 *R gives

1

3

4

5

2 2

5 1 1 1 1

20 20 0 0 0

* 0 0 4 0 0

* 0 0 0 4 0

* 0 0 0 0 4

** 5 *

R

R

R

R

R R

=

Executing the row operation2** 4

1R R+ yields

1

2 2 1

3

4

5

5 1 1 1 1

*** ** 4 0 24 4 4 4

* 0 0 4 0 0

* 0 0 0 4 0

* 0 0 0 0 4

R

R R R

R

R

R

= +

What is the determinant of this matrix?

We have an upper triangular matrix so the determinant is the product of all the entries

on the leading diagonal, that is 5 24 4 4 4 7680 = . What is the determinant of thegiven matrixA?

All the above row operations apart from2

5 *R makes no difference to the determinant.

Why not?

Because by result (2.23) adding a multiple of one row to another has the same

determinant.How does the row operation2

5 *R change the determinant?

(2.23) ( ) ( )det det if a multiple of one row is added to another =B A


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By (2.23) we have the above determinant 7680 is ( )5 det A , that is

( ) ( )7680

5det 7680 which gives det 15365

= =A A =

Proposition (2.24). Let Ebe an elementary matrix. For any square matrix A of the samesize as E we have

( ) ( ) ( )det det det=EA E A


Means the determinant of matrix multiplication EA is equal to the determinant of the

elementary matrix E times the determinant of matrix A.

Proof.

We consider the three different cases of elementary matrices separately.

Case 1. Let Ebe the elementary matrix obtained from the identity matrix by adding a

multiple of one row to another. Then from Chapter 1 we have that the matrix

multiplication EAperforms the same row operation of adding one row to another of

matrix A. By the first line of Result (2.23) we have

( ) ( )det det=EA A

By result (2.21) we have ( )det 1=E therefore

( ) ( )

( ) ( ) ( )

det det

1 det det det

=

= =

EA A

A E A

Hence for the first case we have ( ) ( ) ( )det det det=EA E A .

Case 2. Let Ebe the elementary matrix obtained from the identity matrix by

multiplying a row (or column) by a non-zero scalar k. By result (2.21) we have

( )det k=E . The matrix multiplication EAperforms the same row operation ofmultiplying a row by a non-zero scalar k on matrix A. By result (2.23) we have

( ) (det detk=EA A) .

( )( )

( ) ( ) ( )det

det det det detk=

= =E

EA A E A

We have proven ( ) ( ) ( )det det det=EA E A for the second case.

Case 3. See Exercise 2(b).

Proposition (2.25). Let be elementary matrices and Bbe a1 2 3, , , and

kE E E E

square matrix of the same size. Then

( ) ( ) ( ) ( ) ( ) (1 2 3 1 2 3det det det det det detk k=E E E E B E E E E B )


Means if we have a matrix multiplication then the determinant of this is1 2 3 k

E E E E B

equal to determinant of each matrix multiplied together.

Proof.

Exercise 2(b).

(2.23) ( )

( )

( )

( )




=

A

B A

A


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Next we prove an important propertyof invertible (non-singular) matrices.

Theorem (2.26). A square matrix A is invertible (non-singular) if and only if

( )det 0A


Means if matrix A is invertible then the determinant of A does not equal zero. Also if

the determinant is not equal to zero then the matrix A is invertible (non-singular).

How do we prove this result?

Since it is an if and only if statement, we need to prove it both ways. That is we first

assume matrix A is invertible and show that this leads to ( )det 0A . This direction is

normally symbolized by . Then we assume ( )det 0A and prove that matrix A is

invertible. This direction is normally symbolized by .Proof.

( ). Assume the matrix A is invertible. By Proposition (1.27) part d) of chapter 1 weknow the matrix A is a product of elementary matrices. We can write the matrix A as

1 2 3 k=A E E E E

where are elementary matrices. By Proposition (2.25) we1 2 3, , , and

kE E E E

have

( ) ( )

( ) ( ) ( ) (1 2 3

1 2 3

det det

det det det det

k

k

=

=

A E E E E

E E E E

)

Remember

(2.21) ( )

1 if a multiple of one row is added to another

det 1 if two rows have been interchanged

if a row has been multiplied by non-zerok k

=

E

The determinant of an elementary matrix can only be 1, 1 or the non-zero k.

Multiplying these non-zero numbers, ( ) ( ) ( ) ( )1 2 3det det det det kE E E E , cannot give

0. Therefore .( )det 0A

Now we go the other way ( ). Assume ( )det 0A then by Proposition (2.13) we

have( )

( )11

detadj =A

AA which means that matrix A is invertible (non-singular).

Hence we have proven our result.

We can extend Proposition (2.24) to any two square matrices of the same size as the

next proposition states.

Proposition (2.27). If A and B are square matrices of the same size then

( ) ( ) ( )det det det=AB A B

Proof.

We consider the two cases of matrix A. Case 1 is where the matrix A is invertible (non-

singular) and case 2 is where matrix A is non-invertible (singular).

Case 1. Assume the matrix A is invertible. Then by Proposition (1.27) part d) of chapter

1 we know matrix A is a product of elementary matrices. We can write

1 2 3 k

=A E E E E

(1.27) Part (d) Ais invertible Ais a product of elementary matrices


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where are elementary matrices. We have1 2 3, , , and

kE E E E

( ) ( )

( ) ( ) ( ) ( ) ( ) [ ]

( )

( ) ( )

1 2 3

1 2 3

1 2 3

det det

det det det det det By (2.25)

det det

det det

k

k

k

=

=

=

=

=

A

AB E E E E B

E E E E B

E E E E B

A B

Case 2. Assume matrix A is non-invertible (singular). By the above Proposition (2.26)

we conclude that . Since matrix A is non-invertible therefore matrix( )det 0=A

multiplication AB is also non-invertible. Why?

Suppose AB is invertible then by Proposition (1.20) of the last chapter we have matrix

A (and B)is invertible. Since AB is also non-invertible therefore . Hence( )det 0=AB

we have our result

( ) ( ) ( )det det det=AB A B because and( )det 0=A ( )det 0=AB .

Again we can extend the result of Proposition (2.27) to nsquare matrices as the next

proposition states.

Proposition (2.28). If are square matrices of the same size1 2 3

,, , and n

A A A A

then

( ) ( ) ( ) ( ) ( )1 2 3 1 2 3det det det det detn n=A A A A A A A A

Proof.

Exercise 2(b).

Generally a function in mathematics which has these properties of Proposition (2.27) or(2.28) is called a multiplicative function.

Generally in mathematics we say a functionf is multiplicative if

( ) ( ) ( )f xy f x f y=

The determinant is an example of a multiplicative function.

Proposition (2.29). If A is an invertible (non-singular) matrix then

( )( )

1 1detdet

=AA

Proof. Exercise 2(b)

(1.20) ( ) 1 1 1 =AB B A

(2.25) ( ) ( ) ( ) ( ) ( ) (1 2 3 1 2 3det det det det det detk k=E E E E B E E E E B )

(2.26) A square matrix A is invertible (non-singular) if and only if ( )det 0A


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SUMMARY

A triangular matrix is a square matrix where all entries to oneside of the leading

diagonal are zero.

A diagonal matrix is a square matrix where all entries to bothsides of the leading

diagonal are zero.

The determinant of a triangular or diagonal matrix is a product of the entries along theleading diagonal.

(2.23) ( )

( )

( )

( )




=

A

B A

A

A square matrix A is invertible if and only if the determinant of A does not equal to

zero.

If A and B are square matrices of the same size then

( ) ( ) ( )det det det=AB A B

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Properties of a Determinant