Mathematics

Session

Matrices and Determinants - 2

Session Objectives

Determinant of a Square Matrix

Minors and Cofactors

Properties of Determinants

Applications of Determinants

Area of a TriangleSolution of System of Linear Equations (Cramer’s Rule)

Class Exercise

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

Example

2 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

Minors

4 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+ jij 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 of Determinants

1. The value of a determinant remains unchanged, if its rows and columns are interchanged.

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. . '

2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign.

1 1 1 2 2 2

2 2 2 1 1 1 2 1

3 3 3 3 3 3

a b c a b c

a b c = - a b c R R

a b c a b c

Applying

Properties (Con.)

3. 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.

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

ka kb kc a b c

a b c = k a b c

a b c a b c

which also implies

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a b c ma mb mc1

a b c = a b cm

a b c a b c

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

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.)

6. If any two rows (or columns) of a determinant are identical, then its value is zero.

2 2 2

3 3 3

0 0 0

a b c =0

a b c

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

1 1 1

2 2 2

1 1 1

a b c

a b c =0

a b c

Properties (Con.)

a 0 0

8 Let A = 0 b 0 be a diagonal matrix, then

0 0 c

a 0 0

= 0 b 0

0 0 c

A abc

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 + Rj to denote the addition of times the elements of jth row to the corresponding elements of ith row.

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

Evaluation of Determinants

If 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 that3

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 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+CProve 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

1 1 2 3

2(a+b+c) 2(a+b+c) 2(a+b+c)

= c+a a+b b+c Applying R R +R +R

a+b b+c c+a

1 1 1

=2(a+b+c) c+a a+b b+c

a+b b+c c+a

Example -7

Solution :

Using properties of determinants, prove that

2 2 2

b+c c+a a+b

c+a a+b b+c =2(a+b+c)(ab+bc+ca- a - b - c ).

a+b b+c c+a

b+c c+a a+b

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

a+b b+c c+a

1 1 2 2 2 3

0 0 1

=2(a+b+c) (c- b) (a- c) b+c Applying C C - C and C C - C

(a- c) (b- a) c+a

Now expanding along R1 , we get

22(a+b+c) (c- b)(b- a) - (a- c)

2 2 2=2(a+b+c) bc- b - ac+ab- (a +c - 2ac)

Solution Cont.

2 2 2=2(a+b+c) ab+bc+ac- a -b - c

=R.H.S

Using properties of determinants prove that

2

x+4 2x 2x

2x x+4 2x =(5x+4)(4- x)

2x 2x x+4

Example - 8

1 2x 2x

=(5x+4) 1 x+4 2x

1 2x x+4

Solution :

1 1 2 3

x+4 2x 2x 5x+4 2x 2x

L.H.S = 2x x+4 2x =5x+4 x+4 2x Applying C C +C +C

2x 2x x+4 5x+4 2x x+4

Solution Cont.

2 2 1 3 3 2

1 2x 2x

=(5x+4) 0 -(x - 4) 0 ApplyingR R - R and R R - R

0 x- 4 -(x - 4)

Now expanding along C1 , we get

2(5x+4) 1(x - 4) - 0

2=(5x+4)(4- x)

=R.H.S

Example -9

Using properties of determinants, prove that

x+9 x x

x x+9 x =243 (x+3)

x x x+9

x+9 x x

L.H.S= x x+9 x

x x x+9

1 1 2 3

3x+9 x x

= 3x+9 x+9 x Applying C C +C +C

3x+9 x x+9

Solution :

1=3(x+3) 81 Expanding along C

=243(x+3)

=R.H.S.

1 x x

=(3x+9) 1 x+9 x

1 x x+9

Solution Cont.

2 2 1 3 3 2

1 x x

=3 x+3 0 9 0 Applying R R - R and R R - R

0 -9 9

Example -10

Solution :

2 2 2 2 2

2 2 2 2 21 1 3

2 2 2 2 2

(b+c) a bc b +c a bc

L.H.S.= (c+a) b ca = c +a b ca Applying C C - 2C

(a+b) c ab a +b c ab

2 2 2 2

2 2 2 21 1 2

2 2 2 2

a +b +c a bc

a +b +c b ca Applying C C +C

a +b +c c ab

2

2 2 2 2

2

1 a bc

=(a +b +c ) 1 b ca

1 c ab

2 2

2 2 2 2 2

2 2

(b+c) a bc

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

(a+b) c ab

Show that

Solution Cont.

2

2 2 22 2 1 3 3 2

1 a bc

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

0 (c-b)(c+b) a(b- c)

2 2 2 2 21=(a +b +c )(a- b)(b- c)(-ab- a +bc+c ) Expanding along C

2 2 2=(a +b +c )(a- b)(b- c)(c- a)(a+b+c)=R.H.S.

2

2 2 2

1 a bc

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

0 -(b+c) a

2 2 2=(a +b +c )(a- b)(b- c) b c- a + c- a c+a

Applications of Determinants (Area of a Triangle)

The area of a triangle whose vertices are

is given by the expression

1 1 2 2 3 3(x , y ), (x , y ) and (x , y )

1 1

2 2

3 3

x y 11

Δ = x y 12

x y 1

1 2 3 2 3 1 3 1 2

1= [x (y - y ) + x (y - y ) + x (y - y )]

2

Example

Find the area of a triangle whose vertices are (-1, 8), (-2, -3) and (3, 2).

Solution :

1 1

2 2

3 3

x y 1 -1 8 11 1

Area of triangle= x y 1 = -2 -3 12 2

x y 1 3 2 1

1= -1(-3- 2) - 8(-2- 3)+1(-4+9)

2

1= 5+40+5 =25 sq.units

2

Condition of Collinearity of Three Points

If are three points,

then A, B, C are collinear

1 1 2 2 3 3A (x , y ), B (x , y ) and C (x , y )

1 1 1 1

2 2 2 2

3 3 3 3

Area of triangle ABC =0

x y 1 x y 11

x y 1 =0 x y 1 =02

x y 1 x y 1

If the points (x, -2) , (5, 2), (8, 8) are collinear, find x , using determinants.

Example

Solution :

x -2 1

5 2 1 =0

8 8 1

x 2- 8 - -2 5- 8 +1 40-16 =0

-6x - 6+24=0

6x =18 x =3

Since the given points are collinear.

Solution of System of 2 Linear Equations (Cramer’s Rule)

Let the system of linear equations be

2 2 2a x +b y = c ... ii

1 1 1a x +b y = c ... i

1 2D DThen x = , y = provided D 0,

D D

1 1 1 1 1 11 2

2 2 2 2 2 2

a b c b a cwhere D = , D = and D =

a b c b a c

Cramer’s Rule (Con.)

then the system is consistent and has infinitely many solutions.

1 22 If D = 0 and D =D = 0,

then the system is inconsistent and has no solution.

1 If D 0

Note :

,

then the system is consistent and has unique solution.

1 23 If D = 0 and one of D , D 0,

Example

2 -3D= =2+9=11 0

3 1

1

7 -3D = =7+15=22

5 1

2

2 7D = =10- 21=-11

3 5

Solution :

1 2

D 0

D D22 -11By Cramer's Rule x= = =2 and y= = =-1

D 11 D 11

Using Cramer's rule , solve the following system of equations 2x-3y=7, 3x+y=5

Solution of System of 3 Linear Equations (Cramer’s Rule)

Let the system of linear equations be

2 2 2 2a x +b y +c z = d ... ii

1 1 1 1a x +b y +c z = d ... i

3 3 3 3a x +b y +c z = d ... iii

31 2 DD DThen x = , y = z = provided D 0,

D D D,

1 1 1 1 1 1 1 1 1

2 2 2 1 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3

a b c d b c a d c

where D = a b c , D = d b c , D = a d c

a b c d b c a d c

1 1 1

3 2 2 2

3 3 3

a b d

and D = a b d

a b d

Cramer’s Rule (Con.)

Note:

(1) If D 0, then the system is consistent and has a unique solution.

(2) If D=0 and D1 = D2 = D3 = 0, then the system has infinite solutions or no solution.

(3) If D = 0 and one of D1, D2, D3 0, then the system is inconsistent and has no solution.

(4) If d1 = d2 = d3 = 0, then the system is called the system of homogeneous linear equations.

(i) If D 0, then the system has only trivial solution x = y = z = 0.

(ii) If D = 0, then the system has infinite solutions.

Example

Using Cramer's rule , solve the following system of equations5x - y+ 4z = 5 2x + 3y+ 5z = 25x - 2y + 6z = -1

Solution :

5 -1 4

D= 2 3 5

5 -2 6

1

5 -1 4

D = 2 3 5

-1 -2 6

= 5(18+10)+1(12+5)+4(-4 +3)= 140 +17 –4= 153

= 5(18+10) + 1(12-25)+4(-4 -15)= 140 –13 –76 =140 - 89= 51 0

3

5 -1 5

D = 2 3 2

5 -2 -1

= 5(-3 +4)+1(-2 - 10)+5(-4-15)= 5 – 12 – 95 = 5 - 107= - 102

Solution Cont.

1 2

3

D 0

D D153 102By Cramer's Rule x= = =3, y= = =2

D 51 D 51D -102

and z= = =-2D 51

2

5 5 4

D = 2 2 5

5 -1 6

= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)= 85 + 65 – 48 = 150 - 48= 102

Example

Solve the following system of homogeneous linear equations:

x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0

Solution:

1 1 - 1

We have D = 1 - 2 1 = 1 10 - 6 - 1 -5 - 3 - 1 6 + 6

3 6 - 5

= 4 + 8 - 12 = 0

The system has infinitely many solutions.

Putting z = k, in first two equations, we get

x + y = k, x – 2y = -k

Solution (Con.)

1

k 1

D -k - 2 -2k + k kBy Cramer's rule x = = = =

D -2 - 1 31 1

1 - 2

2

1 k

D 1 - k -k - k 2ky = = = =

D -2 - 1 31 1

1 - 2

k 2kx = , y = , z = k , where k R

3 3

These values of x, y and z = k satisfy (iii) equation.

Thank you