Determinants

Determinants

If is a square matrix of order 1,

then |A| = | a11 | = a11

ijA = a

If is a square matrix of order 2, then11 12

21 22

a aA =

a a

|A| = = a11a22 – a21a12

a a

a a

11 12

21 22

Example

4 - 3Evaluate the determinant :

2 5

4 - 3Solution : = 4 ×5 - 2 × -3 = 20 + 6 = 26

2 5

Solution

If A = is a square matrix of order 3, then11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

[Expanding along first row]

11 12 1322 23 21 23 21 22

21 22 23 11 12 1332 33 31 33 31 32

31 32 33

a a aa a a a a a

| A | = a a a = a - a + aa a a a a a

a a a

11 22 33 32 23 12 21 33 31 23 13 21 32 31 22= a a a - a a - a a a - a a + a a a - a a

11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 22a a a a a a a a a a a a a a a a a a

Example2 3 - 5

Evaluate the determinant : 7 1 - 2

-3 4 1

2 3 - 5

1 - 2 7 - 2 7 17 1 - 2 = 2 - 3 + -5

4 1 -3 1 -3 4-3 4 1

= 2 1 + 8 - 3 7 - 6 - 5 28 + 3

= 18 - 3 - 155

= -140

[Expanding along first row]

Solution :

Minors

-1 4If A = , then

2 3

21 21 22 22M = Minor of a = 4, M = Minor of a = -1

11 11 12 12M = Minor of a = 3, M = Minor of a = 2

Minors4 7 8

If A = -9 0 0 , then

2 3 4

M11 = Minor of a11 = determinant of the order 2 × 2 square sub-matrix is obtained by leaving first row and first column of A

0 0= =0

3 4

Similarly, M23 = Minor of a23

4 7= =12- 14=-2

2 3

M32 = Minor of a32 etc.4 8

= =0+72=72-9 0

Cofactors i+ j

ij ij ijC = Cofactor of a in A = -1 M ,

ij ijwhere M is minor of a in A

Cofactors (Con.)

C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1 0 0

=03 4

C23 = Cofactor of a23 = (–1)2 + 3 M23 = 4 7

22 3

C32 = Cofactor of a32 = (–1)3 + 2M32 = etc.4 8

- =- 72-9 0

4 7 8

A = -9 0 0

2 3 4

Value of Determinant in Terms of Minors and Cofactors

11 12 13

21 22 23

31 32 33

a a a

If A = a a a , then

a a a

3 3

i jij ij ij ij

j 1 j 1

A 1 a M a C

i1 i1 i2 i2 i3 i3= a C +a C +a C , for i =1 or i = 2 or i = 3

Properties

1. If each element of a row (or column) of a determinant is zero, then its value is zero.

2 2 2

3 3 3

0 0 0

a b c =0

a b c

Proof: Expanding the determinant along the row containing only zeros:

0

..21

0..002....2221

1...1211

nnanana

naaanaaa

.00det1

n

jijCA

Properties 2. If all the elements of a row (or column) is multiplied by a

non-zero number k, then the value of the new determinant is k times the value of the original determinant.

)det(...

...

...

...

...

...

)det(

21

21

21

22221

11211

21

ln2211

21

22221

11211

A

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

A

nnnn

knkk

knkk

n

n

nnnn

klkl

knkk

n

n

Proof: expand elong the row multiplied by

n

jijij

n

jijij CaCa

11

detdet AB

Properties (Con.)

4. If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two (or more) determinants.

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3

a +x b c a b c x b c

a +y b c = a b c + y b c

a +z b c a b c z b c

nnananancccnaaanaaa

nnanananbbbnaaanaaa

nnanana

ncnbcbcbnaaanaaa

..21

..21

2....2221

1...1211

..21

..21

2....2221

1...1211

..21

..2211

2....2221

1...1211

Proof: expand along the row containing the sum:

n

j

n

jijijijij

n

jijijij DbDcDcb

1 11

detA

Properties of Determinants2. If any two rows (or columns) of a determinant are interchanged,

then the value of the determinant is changed by minus sign. Proof: first we proove it for neighbourh rows:

nnnnn

ni

in

i

i

i

i

i

i

n

n

aaaa

a

a

a

a

a

a

a

a

aaaa

aaaa

D

321

)1(3)1(

3

2)1(

2

1)1(

1

2232221

1131211

1

.

.

nnnnn

in

ni

i

i

i

i

i

i

n

n

aaaa

a

a

a

a

a

a

a

a

aaaa

aaaa

D

321

)1(

3

3)1(

2

2)1(

1

1)1(

2232221

1131211

2

.

.

n

jjiij DaD

111

n

jijij DaD

12 1

21 -DD

Cont. Interchanging 2 rows (columns)

Suppose, that we change the i. and j. rows, and j-i = k.i i ji+1 j ii+2 i+1 i+1…. i+2 i+2I+k=j …. …

Then we need k neighbourh change to position the j. row right below the i. (see 2nd column) One more change and j stands at the earlier position of i, and i is the row right after j. Again, i must be interchanged by its neighbours k times, so the sum of neighbourgh row changes is: k+k+1=2k+1 – an odd number, so because by each change the deterimant changes it sing, after odd number of changes, its changes its sign.

Properties6. If any two rows (or columns) of a determinant are identical,

then its value is zero.

Proof: if we interchanged the identical rows, the determinant

remains the same, but formally it does changes its sign,

so :D = ( - D ), which is possible only if D=0

5. The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column).

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 1 1 2 3

3 3 3 3 3 3 3 3

a b c a +mb - nc b c

a b c = a +mb - nc b c C C + mC - nC

a b c a +mb - nc b c

Applying

Properties (Con.)

nnnn

jnjj

jninjiji

n

nnnn

jnjj

inii

n

aaa

aaa

aaaaaa

aaa

D

aaa

aaa

aaa

aaa

D

21

21

2211

11211

21

21

21

11211

.

Proof:0

0

21

21

21

11211

21

21

21

11211

D

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

D

nnnn

jnjj

jnjj

n

nnnn

jnjj

inii

n

Properties (Con.)

nn

nnnnnnnn

aaaa

aaa

aa

a

aa

aaa

aa

a

a .........

.

.

0

....

00

.

.

0

....

00

nna..

n2a

n1a

0....

0022

a21

a

00011

a

332211

21

4443

33

2211

21

3332

22

11

If A is an upper (lower) triangular matrix, then the determinant is equal to the product of the elements of the main diagonal:

Proof:

1. The value of a determinant remains unchanged, if its rows and columns are interchanged. (A matrix and its transpose have the same detereminant)

1 1 1 1 2 3

2 2 2 1 2 3

3 3 3 1 2 3

a b c a a a

a b c = b b b

a b c c c c

i e A A. . '

Row(Column) Operations

Following are the notations to evaluate a determinant:

Similar notations can be used to denote column operations by replacing R with C.

(i) Ri to denote ith row

(ii) Ri Rj to denote the interchange of ith and jth rows.

(iii) Ri Ri + lRj to denote the addition of l times the elements of jth row to the corresponding elements of ith row.

(iv) lRi to denote the multiplication of all elements of ith row by l.

Evaluation of DeterminantsIf a determinant becomes zero on putting is the factor of the determinant. x = , then x -

2

3

x 5 2

For example, if Δ = x 9 4 , then at x =2

x 16 8

, because C1 and C2 are identical at x = 2

Hence, (x – 2) is a factor of determinant .

0

Sign System for Expansion of Determinant

Sign System for order 2 and order 3 are given by

+ – ++ –

, – + –– +

+ – +

42 1 6 6×7 1 6

i 28 7 4 = 4×7 7 4

14 3 2 2×7 3 2

1

6 1 6

=7 4 7 4 Taking out 7 common from C

2 3 2

Example-1

6 -3 2

2 -1 2

-10 5 2

42 1 6

28 7 4

14 3 2

Find the value of the following determinants

(i) (ii)

Solution :

1 3= 7×0 C and C are identical

= 0

Example –1 (ii)6 -3 2

2 -1 2

-10 5 2

(ii)

3 2 3 2

1 2 1 2

5 2 5 2

1

1 2

3 3 2

( 2) 1 1 2 Taking out 2 common from C

5 5 2

( 2) 0 C and C are identical

0

Evaluate the determinant1 a b+c1 b c+a1 c a+b

Solution :

3 2 3

1 a b+c 1 a a+b+c

1 b c+a = 1 b a+b+c Applying c c +c

1 c a+b 1 c a+b+c

3

1 a 1

= a+b+c 1 b 1 Taking a+b+c common from C

1 c 1

Example - 2

1 3= a+ b + c ×0 C and C are identical

= 0

2 2 2

a b c

We have a b c

bc ca ab

21 1 2 2 2 3

(a-b) b- c c

= (a-b)(a+b) (b- c)(b+c) c Applying C C - C and C C - C

-c(a-b) -a(b- c) ab

2

1 2

1 1 cTaking a- b and b- c common

=(a- b)(b- c) a+b b+c cfrom C and C respectively

-c -a ab

Example - 3

bc

2 2 2

a b c

a b c

ca ab

Evaluate the determinant:

Solution:

21 1 2

0 1 c

=(a- b)(b- c) -(c- a) b+c c Applying c c - c

-(c- a) -a ab

2

0 1 c

=-(a- b)(b- c)(c- a) 1 b+c c

1 -a ab

22 2 3

0 1 c

= - (a- b)(b- c)(c- a) 0 a+b+c c - ab Applying R R - R

1 -a ab

Now expanding along C1 , we get(a-b) (b-c) (c-a) [- (c2 – ab – ac – bc – c2)]= (a-b) (b-c) (c-a) (ab + bc + ac)

Solution Cont.

Without expanding the determinant,

prove that 3

3x+y 2x x

4x+3y 3x 3x =x

5x+6y 4x 6x

3x+y 2x x 3x 2x x y 2x x

L.H.S= 4x+3y 3x 3x = 4x 3x 3x + 3y 3x 3x

5x+6y 4x 6x 5x 4x 6x 6y 4x 6x

3 2

3 2 1 1 2 1

=x 4 3 3 +x y 3 3 3

5 4 6 6 4 6

Example-4

Solution :

3 21 2

3 2 1

=x 4 3 3 +x y×0 C and C are identical in I I determinant

5 4 6

Solution Cont.

31 1 2

1 2 1

=x 1 3 3 Applying C C - C

1 4 6

32 2 1 3 3 2

1 2 1

=x 0 1 2 ApplyingR R - R and R R - R

0 1 3

31

3

= x ×(3- 2) Expanding along C

=x =R.H.S.

3

3 2 1

=x 4 3 3

5 4 6

Prove that : = 0 , where w is cube root of unity.

3 5

3 4

5 5

1 ω ω

ω 1 ω

ω ω 1

3 5 3 3 2

3 4 3 3

5 5 3 2 3 2

1 ω ω 1 ω ω .ω

L.H.S = ω 1 ω = ω 1 ω .ω

ω ω 1 ω .ω ω .ω 1

2

3

2 2

1 2

1 1 ω

= 1 1 ω ω =1

ω ω 1

=0=R.H.S. C and C are identical

Example -5

Solution :

Example-6

2

x+a b c

a x+b c =x (x+a+b+c)

a b x+C

Prove that :

1 1 2 3

x+a b c x+a+b+c b c

L.H.S= a x+b c = x+a+b+c x+b c

a b x+C x+a+b+c b x+c

Applying C C +C +C

Solution :

1

1 b c

= x+a+b+c 1 x+b c

1 b x+c

Taking x+a+b+c commonfrom C

Solution cont.

2 2 1 3 3 1

1 b c

=(x+a+b+c) 0 x 0

0 0 x

Applying R R -R and R R -R

Expanding along C1 , we get

(x + a + b + c) [1(x2)] = x2 (x + a + b + c)

= R.H.S