Vector Calculus And Linear Algebra

Guided by :Dr. Ravi TailorPatel Vinita .G.

Presented by :Abdul Sattar(13094010700

4)

CRAMER'S RULE

(Using Determinants to solve systems of equations)

INTRODUCTION

Suppose that we have a square system with n equation in the same number of variables ( ). Then the solution of the system has the following cases.

1) If the system has non-zero coefficient determinant D=det(A),

then the system has unique solution and this solution is of the form

where is the determinant obtained from D by replacing in D

column by the column with the entries ( )

nxxx ,......., 21

,,.......,,D

Dx

D

Dx

D

Dx n

n 22

11

iD

thinbbb ,.....,, 21

a) if at least one of Di is non-zero then the system

has no solution b) if all Di‘s are zero, then the system has infinite

number of solutions.

In this case, if the given system is homogeneous; that is, right

hand side is zero then we have ffollowing possibilities of its solution.

(1) if Di≠ 0, then the system has only trival solution.

(2) if D=0, then the system has also nontrival solutions.

2) If the system has zero coefficient determinant D=det(A) , then we have two possibilities as discussed below.

|D|

D = 0Di=0(i=1

,2,3) Infinite solution

Di ≠ 0 (i=1,2,3

) No solutio

n

D ≠ 0Unique solution

EXAMPLE : 1

Find the solution of the system

Solution :In matrix form, the given system of equations can be written as Ax=b, where

74

63

52

zyx

zyx

zyx

7

6

5

,,

411

113

121

b

z

y

x

xA

Here , matrix A is a square matrix of order 3, so Cramer’s rule can be applied Now,

411

113

121

||)det( AAD

)4(1)11(2)5(1 023

)13(1)112(2)14(1

Therefore, the system has unique solution.For finding unique solution, let us first find D1,D2 and D3 it can be easily verified that

)76(1)724(2)14(5

417

116

125

1 D

46

)13(1)17(2)5(5

EXAMPLE : 2

Find the solution of the system

7242

532

32

zyx

zyx

zyx

Solution :In matrix form, the given system of equations can be written as Ax=b, where

7

5

3

,,

242

312

121

b

z

y

x

xA

)621(1)112(5)724(1

471

163

151

2 D

23

)15()11(517

)13(5)621(2)67(1

711

613

521

3 D

23

)4(5)15(213

Therefore, the system has unique solution.

123

23,1

23

23,2

23

46 321

D

Dz

D

Dy

D

Dx

Here , matrix A is a square matrix of order 3, so Cramer’s rule can be applied Now,

)28(1)64(2)122(1

242

312

121

|| AD

6)2(210 0

Therefore, either the system has no solution or infinite number of solution. Let us check for it.

)720(1)2110(2)122(3

247

315

123

1 D

13)11(2)10(3

05Therefore, the system has no solution as at least one Di, i=1,2,3 is nonzero.

EXAMPLE : 3

Find the solution of the system

24987

15654

632

zyx

zyx

zyx

Solution :In matrix form, the given system of equations can be written as Ax=b, where

24

15

6

,,

987

654

321

b

z

y

x

xA

Here , matrix A is a square matrix of order 3, so Cramer’s rule can be applied Now,

)3532(3)4236(2)4845(1

987

654

321

|| AD

)3(3)6(23 0

Also,

)120120(3)144135(2)4845(6

9824

6515

326

1 D

0

)0(3)9(2)3(6

)10596(3)4236(6)144135(1

9247

6154

361

2 D

0

)9(3)6(69

)3532(6)10596(2)120120(1

2487

1554

621

3 D

0

)3(6)9(20

Therefore, the system has infinitle number of solutions.Now, 0385

54

21

Therefore, þ(A)=2Omitting m-r = 3-2 = 1 Considering n-r = 3-2 = 1 variable as arbitary, the remaining system becomes

xzy

xzy

41565

632

Where x is arbitaryNow ,

3151265

32D

96)415(3)636(6415

361

xxxx

xD

xxxx

xD 3)6(5)415(2

4155

621

Therefore ,

xx

D

Dzx

x

D

Dy

3

3,23

3

96 21

Let x = k , where k is arbitary , then the infinite number of solutions of the given system is kzkykx ,23,

where k is an arbitary constant.

THANK YOU !