matrices & determinants chapter: 1 matrices & determinants

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Matrices & Determinants Chapter: 1 Matrices & Determinants

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Page 1: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Chapter: 1 Matrices & Determinants

Page 2: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Session Objectives

• Meaning of matrix• Type of matrices• Transpose of Matrix• Meaning of symmetric and skew symmetric

matrices• Minor & co-factors• Computation of adjoint and inverse of a

matrix

Page 3: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

TYPES OF MATRICESNAME DESCRIPTION EXAMPLE

Rectangular matrix

No. of rows is not equal to no. of columns

Square matrix No. of rows is equal to no. of columns

Diagonal matrix

Non-zero element in principal diagonal and zero in all other positions

Scalar matrix Diagonal matrix in which all the elements on principal diagonal and same

502

126

2 1 3

2 0 1

1 2 4

700

040

002

400

040

004

Page 4: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

TYPES OF MATRICES

NAME DESCRIPTION EXAMPLE

Row matrix A matrix with only 1 row

Column matrix A matrix with only I column

Identity matrix Diagonal matrix having each diagonal element equal to one (I)

Zero matrix A matrix with all zero entries

3 2 1 4

2

3

2 4

1 7

0 0

0 0

Page 5: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

TYPES OF MATRICES

NAME DESCRIPTION EXAMPLE

Upper Triangular matrix

Square matrix having all the entries zero below the principal diagonal

Lower Triangular matrix

Square matrix having all the entries zero above the principal diagonal

700

640

352

736

045

002

Page 6: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & 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

Page 7: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Example

4 - 3Evaluate the determinant :

2 5

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

2 5

Page 8: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 9: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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 :

Page 10: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 11: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 12: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Cofactors

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

ij ijwhere M is minor of a in A

Page 13: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 14: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 15: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 16: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 17: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 18: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 19: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 20: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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 .

Page 21: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 22: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Sign System for Expansion of Determinant

Sign System for order 2 and order 3 are given by

+ – ++ –

, – + –– +

+ – +

Page 23: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 24: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 25: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 26: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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:

Page 27: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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.

Page 28: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 29: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 30: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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 :

Page 31: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 32: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 33: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 34: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 35: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 36: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 37: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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 :

Page 38: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 39: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 40: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 41: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 42: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 43: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 44: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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.

Page 45: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 46: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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,

Page 47: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 48: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 49: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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.

Page 50: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 51: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 52: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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

Page 53: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

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.

Page 54: Matrices & Determinants Chapter: 1 Matrices & Determinants

Matrices & DeterminantsMatrices & Determinants

Thank you