Class 12: Maths Chapter 8 solutions. Complete Class 12 Maths Chapter 8 Notes.

RD Sharma Solutions for Class 12 Maths Chapter 8–Solution of Simultaneous Linear Equations

RD Sharma 12th Maths Chapter 8, Class 12 Maths Chapter 8 solutions

Exercise 8.1 Page No: 8.14

1. Solve the following system of equations by matrix method:

(i) 5x + 7y + 2 = 0

4x + 6y + 3 = 0

(ii) 5x + 2y = 3

3x + 2y = 5

(iii) 3x + 4y – 5 = 0

x – y + 3 = 0

(iv) 3x + y = 19

3x – y = 23

(v) 3x + 7y = 4

x + 2y = -1

(vi) 3x + y = 7

5x + 3y = 12

Solution:

(i) Given 5x + 7y + 2 = 0 and 4x + 6y + 3 = 0

Hence, x = 9/2 and y = -7/2

(ii) Given 5x + 2y = 3

3x + 2y = 5

Hence, x = -1 and y = 4

(iii) Given 3x + 4y – 5 = 0

x – y + 3 = 0

Hence, X = 1 Y = – 2

(iv) Given 3x + y = 19

3x – y = 23

(v) Given 3x + 7y = 4

x + 2y = -1

(vi) Given 3x + y = 7

5x + 3y = 12

2. Solve the following system of equations by matrix method:

(i) x + y –z = 3
2x + 3y + z = 10
3x – y – 7z = 1

(ii) x + y + z = 3

2x – y + z = -1

2x + y – 3z = -9

(iii) 6x – 12y + 25z = 4

4x + 15y – 20z = 3

2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14

2x – y + 3z = 4

x + 2y – 3z = 0

(v) (2/x) – (3/y) + (3/z) = 10

(1/x) + (1/y) + (1/z) = 10

(3/x) – (1/y) + (2/z) = 13

(vi) 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8

2y – 3z = 3

x – 2y + 6z = -2

(viii) 2x + y + z = 2

x + 3y – z = 5

3x + y – 2z = 6

(ix) 2x + 6y = 2

3x – z = -8

2x – y + z = -3

(x) 2y – z = 1

x – y + z = 2

2x – y = 0

(xi) 8x + 4y + 3z = 18

2x + y + z = 5

x + 2y + z = 5

(xii) x + y + z = 6

x + 2z = 7

3x + y + z = 12

(xiii) (2/x) + (3/y) + (10/z) = 4,

(4/x) – (6/y) + (5/z) = 1,

(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0

(xiv) x – y + 2z = 7

3x + 4y – 5z = -5

2x – y + 3z = 12

Solution:

(i) Given x + y –z = 3

2x + 3y + z = 10

3x – y – 7z = 1

= (– 20) – 1(– 17) – 1(11)

= – 20 + 17 + 11 = 8

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 21 + 1 = – 20

C21 = (– 1)2 + 1 – 7 – 1 = 8

C31 = (– 1)3 + 1 1 + 3 = 4

C12 = (– 1)1 + 2 – 14 – 3 = 17

C22 = (– 1)2 + 1 – 7 + 3 = – 4

C32 = (– 1)3 + 1 1 + 2 = – 3

C13 = (– 1)1 + 2 – 2 – 9 = – 11

C23 = (– 1)2 + 1 – 1 – 3 = 4

C33 = (– 1)3 + 1 3 – 2 = 1

(ii) Given x + y + z = 3

2x – y + z = -1

2x + y – 3z = -9

= (3 – 1) – 1(– 6 – 2) + 1(2 + 2)

= 2 + 8 + 4

= 14

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 3 – 1 = 2

C21 = (– 1)2 + 1 – 3 – 1 = 4

C31 = (– 1)3 + 1 1 + 1 = 2

C12 = (– 1)1 + 2 – 6 – 2 = 8

C22 = (– 1)2 + 1 – 3 – 2 = – 5

C32 = (– 1)3 + 1 1 – 2 = 1

C13 = (– 1)1 + 2 2 + 2 = 4

C23 = (– 1)2 + 1 1 – 2 = 1

C33 = (– 1)3 + 1 – 1 – 2 = – 3

(iii) Given 6x – 12y + 25z = 4

4x + 15y – 20z = 3

2x + 18y + 15z = 10

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)

= 3510 + 1200 + 1050

= 5760

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (225 + 360) = 585

C21 = (– 1)2 + 1 (– 180 – 450) = 630

C31 = (– 1)3 + 1 (240 – 375) = – 135

C12 = (– 1)1 + 2 (60 + 40) = – 100

C22 = (– 1)2 + 1 (90 – 50) = 40

C32 = (– 1)3 + 1 (– 120 – 100) = 220

C13 = (– 1)1 + 2 (72 – 30) = 42

C23 = (– 1)2 + 1(108 + 24) = – 132

C33 = (– 1)3 + 1 (90 + 48) = 138

(iv) Given 3x + 4y + 7z = 14

2x – y + 3z = 4

x + 2y – 3z = 0

= 3(3 – 6) – 4(– 6 – 3) + 7(4 + 1)

= – 9 + 36 + 35

= 62

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 3 – 6 = – 3

C21 = (– 1)2 + 1 – 12 – 14 = 26

C31 = (– 1)3 + 112 + 7 = 19

C12 = (– 1)1 + 2 – 6 – 3 = 9

C22 = (– 1)2 + 1 – 3 – 7 = – 10

C32 = (– 1)3 + 1 9 – 14 = 5

C13 = (– 1)1 + 2 4 + 1 = 5

C23 = (– 1)2 + 1 6 – 4 = – 2

C33 = (– 1)3 + 1 – 3 – 8 = – 11

(v) Given (2/x) – (3/y) + (3/z) = 10

(1/x) + (1/y) + (1/z) = 10

(3/x) – (1/y) + (2/z) = 13

= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)

= – 10 – 15 + 3

= – 22

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (4 – 6) = – 2

C21 = (– 1)2 + 1(12 – 2) = – 10

C31 = (– 1)3 + 1(9 – 1) = 8

C12 = (– 1)1 + 2 (8 – 3) = – 5

C22 = (– 1)2 + 1 20 – 1 = 19

C32 = (– 1)3 + 1 15 – 2 = – 13

C13 = (– 1)1 + 2 (4 – 2) = 2

C23 = (– 1)2 + 1 10 – 3 = – 7

C33 = (– 1)3 + 1 5 – 6 = – 1

(vi) Given 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

= 3(12 – 6) – 4(0 + 3) + 2(0 – 2)

= 18 – 12 – 4

= 2

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 (12 – 6) = 6

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1(– 12 – 4) = – 16

C12 = (– 1)1 + 2 (0 + 3) = – 3

C22 = (– 1)2 + 1 18 – 2 = 16

C32 = (– 1)3 + 1 – 9 – 0 = 9

C13 = (– 1)1 + 2 (0 – 2) = – 2

C23 = (– 1)2 + 1 (– 6 – 4) = 10

C33 = (– 1)3 + 1 6 – 0 = 6

(vii) Given 3x + 4y + 2z = 8

2y – 3z = 3

x – 2y + 6z = -2

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 6 + 1 = – 5

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1 – 1 – 3 = – 4

C12 = (– 1)1 + 2 – 2 + 3 = – 1

C22 = (– 1)2 + 1 – 4 – 3 = – 7

C32 = (– 1)3 + 1 – 2 – 1 = 3

C13 = (– 1)1 + 21 – 9 = – 8

C23 = (– 1)2 + 12 – 3 = – 1

C33 = (– 1)3 + 1 6 – 1 = 5

(viii) Given 2x + y + z = 2

x + 3y – z = 5

3x + y – 2z = 6

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 – 6 + 1 = – 5

C21 = (– 1)2 + 1(24 + 4) = – 28

C31 = (– 1)3 + 1 – 1 – 3 = – 4

C12 = (– 1)1 + 2 – 2 + 3 = – 1

C22 = (– 1)2 + 1 – 4 – 3 = – 7

C32 = (– 1)3 + 1 – 2 – 1 = 3

C13 = (– 1)1 + 21 – 9 = – 8

C23 = (– 1)2 + 12 – 3 = – 1

C33 = (– 1)3 + 1 6 – 1 = 5

(ix) Given 2x + 6y = 2

3x – z = -8

2x – y + z = -3

= 2(0 – 1) – 6(3 + 2)

= – 2 – 30

= – 32

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 0 – 1 = – 1

C21 = (– 1)2 + 16 + 0 = – 6

C31 = (– 1)3 + 1 – 6 = – 6

C12 = (– 1)1 + 2 3 + 2 = 5

C22 = (– 1)2 + 1 2 – 0 = 2

C32 = (– 1)3 + 1 – 2 – 0 = 2

C13 = (– 1)1 + 2 – 3 – 0 = – 3

C23 = (– 1)2 + 1 – 2 – 12 = 14

C33 = (– 1)3 + 1 0 – 18 = – 18

(x) Given 2y – z = 1

x – y + z = 2

2x – y = 0

= 0 + 4 – 1

= 3

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 1 – 0 = 1

C21 = (– 1)2 + 11 – 2 = 1

C31 = (– 1)3 + 10 + 1 = 1

C12 = (– 1)1 + 2 – 2 – 0 = 2

C22 = (– 1)2 + 1 – 1 – 0 = – 1

C32 = (– 1)3 + 1 0 – 2 = 2

C13 = (– 1)1 + 2 4 – 0 = 4

C23 = (– 1)2 + 1 2 – 0 = – 2

C33 = (– 1)3 + 1 – 1 + 2 = 1

(xi) Given 8x + 4y + 3z = 18

2x + y + z = 5

x + 2y + z = 5

= 8(– 1) – 4(1) + 3(3)

= – 8 – 4 + 9

= – 3

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 1 – 2 = – 1

C21 = (– 1)2 + 1 4 – 6 = 2

C31 = (– 1)3 + 1 4 – 3 = 1

C12 = (– 1)1 + 2 2 – 1 = – 1

C22 = (– 1)2 + 1 8 – 3 = 5

C32 = (– 1)3 + 1 8 – 6 = – 2

C13 = (– 1)1 + 2 4 – 1 = 3

C23 = (– 1)2 + 1 16 – 4 = – 12

C33 = (– 1)3 + 1 8 – 8 = 0

(xii) Given x + y + z = 6

x + 2z = 7

3x + y + z = 12

= 1(– 2) – 1(1 – 6) + 1(1)

= – 2 + 5 + 1

= 4

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 0 – 2 = – 2

C21 = (– 1)2 + 1 1 – 1 = 0

C31 = (– 1)3 + 1 2 – 0 = 2

C12 = (– 1)1 + 2 1 – 6 = 5

C22 = (– 1)2 + 1 1 – 3 = – 2

C32 = (– 1)3 + 1 2 – 1 = – 1

C13 = (– 1)1 + 2 1 – 0 = 1

C23 = (– 1)2 + 1 1 – 3 = 2

C33 = (– 1)3 + 1 0 – 1 = – 1

(xiii) Given (2/x) + (3/y) + (10/z) = 4,

(4/x) – (6/y) + (5/z) = 1,

(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0

AX = B

Now,

|A| = 2(75) – 3(– 110) + 10(72)

= 150 + 330 + 720

= 1200

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 120 – 45 = 75

C21 = (– 1)2 + 1 – 60 – 90 = 150

C31 = (– 1)3 + 1 15 + 60 = 75

C12 = (– 1)1 + 2 – 80 – 30 = 110

C22 = (– 1)2 + 1 – 40 – 60 = – 100

C32 = (– 1)3 + 1 10 – 40 = 30

C13 = (– 1)1 + 2 36 + 36 = 72

C23 = (– 1)2 + 1 18 – 18 = 0

C33 = (– 1)3 + 1 – 12 – 12 = – 24

(xiv) Given x – y + 2z = 7

3x + 4y – 5z = -5

2x – y + 3z = 12

A X = B

Now,

|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)

= 7 + 19 – 22

= 4

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are

C11 = (– 1)1 + 1 12 – 5 = 7

C21 = (– 1)2 + 1 – 3 + 2 = 1

C31 = (– 1)3 + 1 5 – 8 = – 3

C12 = (– 1)1 + 2 9 + 10 = – 19

C22 = (– 1)2 + 1 3 – 4 = – 1

C32 = (– 1)3 + 1 – 5 – 6 = 11

C13 = (– 1)1 + 2 – 3 – 8 = – 11

C23 = (– 1)2 + 1 – 1 + 2 = – 1

C33 = (– 1)3 + 1 4 + 3 = 7

3. Show that each one of the following systems of linear equations is consistent and also find their solutions:

(i) 6x + 4y = 2

9x + 6y = 3

(ii) 2x + 3y = 5

6x + 9y = 15

(iii) 5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

(v) x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

(vi) 2x + 2y – 2z = 1

4x + 4y – z = 2

6x + 6y + 2z = 3

Solution:

(i) Given 6x + 4y = 2

9x + 6y = 3

|A| = 36 – 36 = 0

So, A is singular, Now X will be consistence if (Adj A) x B = 0

C11 = (– 1)1 + 1 6 = 6

C12 = (– 1)1 + 2 9 = – 9

C21 = (– 1)2 + 1 4 = – 4

C22 = (– 1)2 + 2 6 = 6

(ii) Given 2x + 3y = 5

6x + 9y = 15

|A| = 18 – 18 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11 = (– 1)1 + 1 9 = 9

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 3 = – 3

C22 = (– 1)2 + 2 2 = 2

(iii) Given 5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

|A| = 5(260 – 4) – 3(30 – 14) + 7(6 – 182)

= 5(256) – 3(16) + 7(176)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A) x B≠0 or (Adj A) x B = 0

Cofactors of A are

C11 = (– 1)1 + 1 260 – 4 = 256

C21 = (– 1)2 + 1 30 – 14 = – 16

C31 = (– 1)3 + 1 6 – 182 = – 176

C12 = (– 1)1 + 2 30 – 14 = – 16

C22 = (– 1)2 + 1 50 – 49 = 1

C32 = (– 1)3 + 1 10 – 21 = 11

C13 = (– 1)1 + 2 6 – 182 = – 176

C23 = (– 1)2 + 1 10 – 21 = 11

C33 = (– 1)3 + 1 130 – 9 = 121

Now, AX = B has infinite many solution

Let z = k

Then, 5x + 3y = 4 – 7k

3x + 26y = 9 – 2k

This can be written as

(v) Given x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

|A| = 1(2) – 1(4) + 1(2)

= 2 – 4 + 2

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A) x B≠0 or (Adj A) x B = 0

Cofactors of A are

C11 = (– 1)1 + 1 14 – 12 = 2

C21 = (– 1)2 + 1 7 – 4 = – 3

C31 = (– 1)3 + 1 3 – 2 = 1

C12 = (– 1)1 + 2 7 – 3 = – 4

C22 = (– 1)2 + 1 7 – 1 = 6

C32 = (– 1)3 + 1 3 – 1 = 2

C13 = (– 1)1 + 2 4 – 2 = 2

C23 = (– 1)2 + 1 4 – 1 = – 3

C33 = (– 1)3 + 1 2 – 1 = 1

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 – k

x + 2y = 14 – 3k

This can be written as:

(vi) Given x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

This can be written as

|A| = 2(14) – 2(14) – 2(0)

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent

with infinitely many solution according to as:

(Adj A) x B≠0 or (Adj A) x B = 0

Cofactors of A are:

C11 = (– 1)1 + 18 + 6 = 14

C21 = (– 1)2 + 1 4 + 12 = – 16

C31 = (– 1)3 + 1 – 2 + 8 = 6

C12 = (– 1)1 + 2 8 + 6 = – 14

C22 = (– 1)2 + 1 4 + 12 = 16

C32 = (– 1)3 + 1 – 2 + 8 = – 6

C13 = (– 1)1 + 2 24 – 24 = 0

C23 = (– 1)2 + 1 12 – 12 = 0

C33 = (– 1)3 + 1 8 – 8 = 0

4. Show that each one of the following systems of linear equations is consistent:

(i) 2x + 5y = 7

6x + 15y = 13

(ii) 2x + 3y = 5

6x + 9y = 10

(iii) 4x – 2y = 3

6x – 3y = 5

(iv) 4x – 5y – 2z = 2

5x – 4y + 2z = -2

2x + 2y + 8z = -1

(v) 3x – y – 2z = 2

2y – z = -1

3x – 5y = 3

(vi) x + y – 2z = 5

x – 2y + z = -2

-2x + y + z = 4

Solution:

(i) Given 2x + 5y = 7

6x + 15y = 13

|A| = 30 – 30 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11 = (– 1)1 + 1 15 = 15

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 5 = – 5

C22 = (– 1)2 + 2 2 = 2

(ii) Given 2x + 3y = 5

6x + 9y = 10

|A| = 18 – 18 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11 = (– 1)1 + 1 9 = 9

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 3 = – 3

C22 = (– 1)2 + 2 2 = 2

(iii) Given 4x – 2y = 3

6x – 3y = 5

|A| = – 12 + 12 = 0

So, A is singular,

Now X will be consistence if (Adj A) x B = 0

C11 = (– 1)1 + 1 – 3 = – 3

C12 = (– 1)1 + 2 6 = – 6

C21 = (– 1)2 + 1 – 2 = 2

C22 = (– 1)2 + 2 4 = 4

(iv) Given 4x – 5y – 2z = 2

5x – 4y + 2z = -2

2x + 2y + 8z = -1

|A| = 4(– 36) + 5(36) – 2(18)

|A| = 0

Cofactors of A are:

C11 = (– 1)1 + 1 – 32 – 4 = – 36

C21 = (– 1)2 + 1 – 40 + 4 = – 36

C31 = (– 1)3 + 1 – 10 – 8 = – 18

C12 = (– 1)1 + 2 40 – 4 = – 36

C22 = (– 1)2 + 1 32 + 4 = 36

C32 = (– 1)3 + 1 8 + 10 = – 18

C13 = (– 1)1 + 2 10 + 8 = 18

C23 = (– 1)2 + 1 8 + 10 = – 18

C33 = (– 1)3 + 1 – 16 + 25 = 9

(v) Given 3x – y – 2z = 2

2y – z = -1

3x – 5y = 3

|A| = 3(– 5) + 1(3) – 2(– 6)

|A| = 0

Cofactors of A are

C11 = (– 1)1 + 1 0 – 5 = – 5

C21 = (– 1)2 + 1 0 – 10 = 10

C31 = (– 1)3 + 1 1 + 4 = 5

C12 = (– 1)1 + 2 0 + 3 = – 3

C22 = (– 1)2 + 1 0 + 6 = 6

C32 = (– 1)3 + 1 – 3 – 0 = 3

C13 = (– 1)1 + 2 0 – 6 = – 6

C23 = (– 1)2 + 1 – 15 + 3 = 12

C33 = (– 1)3 + 1 6 – 0 = 6

(vi) Given x + y – 2z = 5

x – 2y + z = -2

-2x + y + z = 4

|A| = 1(– 3) – 1(3) – 2(– 3) = – 3 – 3 + 6

|A| = 0

Cofactors of A are:

C11 = (– 1)1 + 1 – 2 – 1 = – 3

C21 = (– 1)2 + 1 1 + 2 = – 3

C31 = (– 1)3 + 1 1 – 4 = – 3

C12 = (– 1)1 + 2 1 + 2 = – 3

C22 = (– 1)2 + 1 1 – 4 = – 3

C32 = (– 1)3 + 1 1 + 2 = – 3

C13 = (– 1)1 + 2 1 – 4 = – 3

C23 = (– 1)2 + 1 1 + 2 = – 3

C33 = (– 1)3 + 1 – 2 – 1 = – 3

x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7

Solution:

2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3.

Solution:

|A| = 2(0) + 3(– 2) + 5(1)

= – 1

Now, the cofactors of A

C11 = (– 1)1 + 1 – 4 + 4 = 0

C21 = (– 1)2 + 1 6 – 5 = – 1

C31 = (– 1)3 + 1 12 – 10 = 2

C12 = (– 1)1 + 2 – 6 + 4 = 2

C22 = (– 1)2 + 1 – 4 – 5 = – 9

C32 = (– 1)3 + 1 – 8 – 15 = 23

C13 = (– 1)1 + 2 3 – 2 = 1

C23 = (– 1)2 + 1 2 + 3 = – 5

C33 = (– 1)3 + 1 4 + 9 = 13

x + 2y + 5z = 10, x – y – z = -2, 2x + 3y – z = -11.

Solution:

Given

|A| = 1(1 + 3) + 2(– 1 + 2) + 5(3 + 2)

= 4 + 2 + 25

= 27

Now, the cofactors of A

C11 = (– 1)1 + 1 1 + 3 = 4

C21 = (– 1)2 + 1 – 2 – 15 = 17

C31 = (– 1)3 + 1 – 2 + 5 = 3

C12 = (– 1)1 + 2 – 1 + 2 = – 1

C22 = (– 1)2 + 1 – 1 – 10 = – 11

C32 = (– 1)3 + 1 – 1 – 5 = 6

C13 = (– 1)1 + 2 3 + 2 = 5

C23 = (– 1)2 + 1 3 – 4 = 1

C33 = (– 1)3 + 1 – 1 – 2 = – 3

Solution:

Given

|A| = 1(1 + 6) + 2(2 – 0) + 0

= 7 + 4

= 11

Now, the cofactors of A

C11 = (– 1)1 + 1 1 + 6 = 7

C21 = (– 1)2 + 1 – 2 – 0 = 2

C31 = (– 1)3 + 1 – 6 – 0 = – 6

C12 = (– 1)1 + 2 2 – 0 = – 2

C22 = (– 1)2 + 1 1 – 0 = 1

C32 = (– 1)3 + 1 3 – 0 = – 3

C13 = (– 1)1 + 2 – 4 – 0 = – 4

C23 = (– 1)2 + 1 – 2 – 0 = 2

C33 = (– 1)3 + 1 1 + 4 = 5

Solution:

Given

|A| = 3(3 – 0) + 4(2 – 5) + 2(0 – 3)

= 9 – 12 – 6

= – 9

Now, the cofactors of A

C11 = (– 1)1 + 1 3 – 0 = 3

C21 = (– 1)2 + 1 – 4 – 0 = 4

C31 = (– 1)3 + 1 – 20 – 6 = – 26

C12 = (– 1)1 + 2 2 – 5 = 3

C22 = (– 1)2 + 1 3 – 2 = 1

C32 = (– 1)3 + 1 15 – 4 = – 11

C13 = (– 1)1 + 2 0 – 3 = – 3

C23 = (– 1)2 + 1 0 + 4 = – 4

C33 = (– 1)3 + 1 9 + 8 = 17

Solution:

Solution:

Given

|A| = 1(– 1 – 1) – 2(– 2 – 0) + 0

= – 2 + 4

= 2

Now, the cofactors of A

C11 = (– 1)1 + 1 – 1 – 1 = – 2

C21 = (– 1)2 + 1 2 – 0 = 2

C31 = (– 1)3 + 1 – 2 – 0 = – 2

C12 = (– 1)1 + 2 2 – 0 = – 2

C22 = (– 1)2 + 1 1 – 0 = 1

C32 = (– 1)3 + 1 – 1 – 0 = 1

C13 = (– 1)1 + 2 – 2 – 0 = – 2

C23 = (– 1)2 + 1 – 1 – 0 = 1

C33 = (– 1)3 + 1 – 1 + 4 = 3

Solution:

Given

9. The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.

Solution:

Let the numbers are x, y, z

x + y + z = 2
…… (i)

Also, 2y + (x + z) + 1

x + 2y + z = 1 …… (ii)

Again,

x + z + 5(x) = 6

5x + y + z = 6 …… (iii)

A X = B

|A| = 1(1) – 1(– 4) + 1(– 9)

= 1 + 4 – 9

= – 4

Hence, the unique solution given by x = A – 1B

C11 = (– 1)1 + 1 (2 – 1) = 1

C12 = (– 1)1 + 2 (1 – 5) = 4

C13 = (– 1)1 + 3 (1 – 10) = – 9

C21 = (– 1)2 + 1 (1 – 1) = 0

C22 = (– 1)2 + 2 (1 – 5) = – 4

C23 = (– 1)2 + 3 (1 – 5) = 4

C31 = (– 1)3 + 1 (1 – 2) = – 1

C32 = (– 1)3 + 2 (1 – 1) = 0

C33 = (– 1)3 + 3 (2 – 1) = 1

10. An amount of ₹10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined incomes are ₹1310 and the combined income of first and second investment is ₹ 190 short of the income from the third. Find the investment in each using matrix method.

Solution:

Let the numbers are x, y, and z

x + y + z = 10,000 …… (i)

Also,

0.1x + 0.12y + 0.15z = 1310 …… (ii)

Again,

0.1x + 0.12y – 0.15z = – 190 …… (iii)

A X = B

|A| = 1(– 0.036) – 1(– 0.03) + 1(0)

= – 0.006

Hence, the unique solution given by x = A – 1B

C11 = – 0.036

C12 = 0.27

C13 = 0

C21 = 0.27

C22 = – 0.25

C23 = – 0.02

C31 = 0.03

C32 = – 0.05

C33 = 0.02

Exercise 8.2 Page No: 8.20

Solve the following systems of homogeneous linear equations by matrix method:

1. 2x – y + z = 0

3x + 2y – z = 0

x + 4y + 3z = 0

Solution:

Given

2x – y + z = 0

3x + 2y – z = 0

X + 4y + 3z = 0

The system can be written as

A X = 0

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)

|A| = 2(10) + 10 + 10

|A| = 40 ≠ 0

Since, |A|≠ 0, hence x = y = z = 0 is the only solution of this homogeneous equation.

2. 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0

Solution:

Given 2x – y + 2z = 0

5x + 3y – z = 0

X + 5y – 5z = 0

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

3. 3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Solution:

Given 3x – y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z = 0

A X = 0

Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)

|A| = – 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x – y = – 2k

4x + 3y = – 3k

4. x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0

Solution:

Given x + y – 6z = 0

x – y + 2z = 0

– 3x + y + 2z = 0

A X = 0

Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)

|A| = – 4 – 8 + 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = 6k

x – y = – 2k

RD Sharma Solutions for Class 12 Maths Chapter 8: Download PDF

RD Sharma Solutions for Class 12 Maths Chapter 8–Solution of Simultaneous Linear Equations

Download PDF: RD Sharma Solutions for Class 12 Maths Chapter 8–Solution of Simultaneous Linear Equations PDF

Chapterwise RD Sharma Solutions for Class 12 Maths :

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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