# Class 12 RD Sharma Solutions – Chapter 8 Solution of Simultaneous Linear Equations – Exercise 8.1 | Set 1

### 4x + 6y + 3 = 0

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= 30 – 28

= 2

The given system has a unique solution given by, X = A-1 B.

Let Cij be the cofactor of the elements aij in A.

C11 = (-1)1+1 (6) = 6, C12 = (-1)1+2 (4) = -4, C21 = -12+1 (7) = -7 and C22 = (-1)2+2 (5) = 5

A-1

A-1

So, X = A-1 B

=>

Therefore, x = 9/2 and y = -7/2.

### 3x + 2y = 5

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= 10 – 6

= 4

The given system has a unique solution given by, X = A-1 B

Let Cij be the cofactor of the elements aij in A.

C11 = -11+1 (2) = 2, C12 = (-1)1+2 (3) = – 3, C21 = (-1)2+1 (2) = – 2 and C22 = (-1)2+2 (5) = 5

A-1

Now, X = A-1 B

=>

Therefore, x = – 1 and y = 4.

### x âˆ’ y + 3 = 0

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= – 3 – 4

= -7

So, the given system has a unique solution given by, X = A-1

Let Cij be the cofactors of the elements aij in A.

C11 = (-1)1+1 (1) = -1, C12 = (-1)1+2 (-1) = 1, C21 = (-1)2+1 (4) = -4 and C22 = (-1)2+2 (3) = 3

Now, X = A-1 B

=>

Therefore, x = -1 and y = 2.

### 3x âˆ’ y = 23

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= – 3 – 3

= -6

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

C11 = (-1)1+1 (-1) = -1, C12 = (-1)1+2 (3) = -3, C21 = (-1)2+1 (1) = -4 and C22 = (-1)2+2 (3) = 3

Now, X = A-1 B

=>

Therefore, x = 7 and y = -2.

### x + 2y = âˆ’1

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= 6 – 7

= -1

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

C11 = (-1)1+1 (2) = 2, C12 = (-1)1+2 (1) = -1, C21 = (-1)2+1 (7) = -7 and C22 = (-1)2+2 (3) = 3

X = A-1 B

=>

Therefore x = – 15 and y = 7.

### 5x + 3y = 12

Solution:

The given system of equations can be written in matrix form as,

AX = B

Here,

A = , X = and B =

Now,

|A| =

= 9 – 5

= 4

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

C11 = (-1)1+1 (3) = 3, C12 = (-1)1+2 (5) = -5, C21 = (-1)2+1 (1) = -1 and C22 = (-1)2+2 (3) = 3

X = A-1 B

=>

Therefore x = 9/4 and y = 1/4.

### 3x âˆ’ y âˆ’ 7z = 1

Solution:

A =

|A| =

= 1 (- 21 + 1) – 1(-14 – 3) – 1(-2 – 9)

= – 20 + 17 + 11

= 8

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactor of the elements aij in A.

X = A-1 B

=> x = 24/8, y = 8/8 and z = 8/8

Therefore, x = 3, y = 1 and z = 1.

### 2x + y âˆ’ 3z = âˆ’ 9

Solution:

A =

|A| =

= 1 (3 – 1) – 1 (-6 – 2) + 1 (2 + 2)

= 2 + 8 + 4

= 14

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

Now, X = A-1 B

=> x = -16/14, y = 20/14 and z = 38/14

Therefore, x = -8/7, y = 10/7 and z = 19/7.

### 2x + 18y + 15z = 10

Solution:

A =

|A| =

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

= 3510 + 1200 + 1050

= 5760

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

Now, X = A-1 B

=> x = 2880/5760, y = 1920/5760 and z = 1152/5760

Therefore x = 1/2, y = 1/3 and z = 1/5.

### x + 2y âˆ’ 3z = 0

Solution:

A =

|A| =

= 3 (3 – 3) – 4 (- 6 – 6) + 7 (2 + 2)

= 0 + 48 + 28

= 76

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

Now, X = A-1 B

=> x = 76/76, y = 76/76 and z = 76/76

Therefore x = 1, y = 1 and z = 1.

### (v)

Solution:

Let 1/x be a, 1/y be b and 1/z be c.

Here,

A =

|A| =

= 2 (2 + 1) + 3 (2 – 3) + 3 (-1 – 3)

= 6 – 3 – 12

= -9

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 1/a = – 9/-18, y = 1/b = – 9/- 27 and z = 1/c = -9/-45

Therefore x = 1/2, y = 1/3 and z = 1/5.

### x + 2y + 4z = 25

Solution:

A =

|A| =

= 5 (4 – 6) – 3 (8 – 3) + 1 (4 – 1)

= -10 – 15 + 3

= – 22

So, the given system has a unique solution given by X = A-1 B.

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = – 22/- 22, y = – 44/- 22 and z = -110/-22

Therefore x = 1, y = 2 and z = 5.

### x âˆ’ 2y + 6z = âˆ’2

Solution:

A =

|A| =

= 3 (12 – 6) – 4 (0 + 3) + 2 (0 – 2)

= 18 – 12 – 4

= 2

Let Cij be the cofactors of the elements aij in A.

Now X = A-1 B

=> x = -4/2, y = 6/2 and z = 2/2

Therefore x = -2, y = 3 and z = 1.

### 3x + y âˆ’ 2z = 6

Solution:

Here,

A =

|A| =

= – 10 – 1 – 8

= -19

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

x = -19/-19, y = -19/-19 and z = 19/-19

Therefore x = 1, y = 1 and z = – 1.

### 2x âˆ’ y + z = âˆ’3

Solution:

A =

|A| =

= 2 (0 – 1) – 6 (3 + 2) + 0 (-3 + 0)

= -2 – 30

= – 32

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 64/-32, y = -32/-32 and z = -64/-32

Therefore x = – 2, y = 1 and z = 2.

### 2y âˆ’ z = 1

Solution:

A =

|A| =

= 1 (1 – 0) + 1 (-2 – 0) + 1(4 – 0)

= 1 – 2 + 4

= 3

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 3/3, y = 6/3 and z = 9/3

Therefore x = 1, y = 2 and z = 3.

### x + 2y + z = 5

Solution:

A =

|A| =

= 8 (1 – 2) – 4 (2 – 1) + 3(4 – 1)

= – 8 – 4 + 9

= -3

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = -3/-3, y = -3/-3 and z = -6/-3

Therefore x = 1, y = 1 and z = 2.

### 3x + y + z = 12

Solution:

A =

|A| =

= 1 (0 – 2) – 1 (1 – 6) + 1(1 – 0)

= – 2 + 5 + 1

= 4

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 12/4, y = 4/4 and z = -20/4

Therefore x = 3, y = 1 and z = – 5.

### (xiii) , x, y, z â‰  0

Solution:

Let 1/x be a, 1/y be b and 1/z be c.

Here,

A =

|A| =

= 2 (120 – 45) – 3 (-80 – 30) + 10 (36 + 36)

= 150 + 330 + 720

= 1200

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 1/a = 1200/600, y = 1/b = 1200/400 and z = 1/c = 1200/240

Therefore x = 2, y = 3 and z = 5.

### 2x âˆ’ y + 3z = 12

Solution:

A =

|A| =

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

= 7 + 19 – 22

= 4

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = 8/4, y = 4/4 and z = 12/4

Therefore x = 2, y = 1 and z = 3.

### 9x + 6y = 3

Solution:

Here,

6x + 4y = 2

9x + 6y = 3

We know, AX = B

A = , X =  and B =

|A| =

= 36 – 36

= 0

Let Cij be the cofactors of the elements aij in A.

C11 = 6, C12 = -9, C21 = -4 and C22 = 6

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### 6x + 9y = 15

Solution:

Here,

2x + 3y = 5

6x + 9y = 15

We know, AX = B

A = , X =  and B =

|A| =

= 18 – 18

= 0

Let Cij be the cofactors of the elements aij in A.

C11 = 9, C12 = -6, C21 = -3 and C22 = 2

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### 7x + 2y + 10z = 5

Solution:

Here,

5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

We know, AX = B

A = , X =  and B =

|A| =

= 1280 – 48 – 1232\]

= 0

Let Cij be the cofactors of the elements aij in A.

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### âˆ’x âˆ’2y + 2z = 1

Solution:

Here,

x âˆ’ y + z = 3

2x + y âˆ’ z = 2

âˆ’x âˆ’2y + 2z = 1

We know, AX = B

A = , X =  and B =

|A| =

= 1\left( 2 – 2 \right) + 1\left( 4 – 1 \right) + 1( – 4 + 1)\]

= 0 + 3 – 3

= 0

Let Cij be the cofactors of the elements aij in A.

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### x + 4y + 7z = 30

Solution:

Here,

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

A = , X =  and B =

|A| =

= 2 – 4 + 2

= 0

Let Cij be the cofactors of the elements aij in A.

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### 6x + 6y + 2z = 3

Solution:

Here,

2x + 2y âˆ’ 2z = 1

4x + 4y âˆ’ z = 2

6x + 6y + 2z = 3

A = , X =  and B =

|A| =

= 2 (8 + 6) – 2 (8 + 6) – 2 (24 – 24)

= 28 – 28 – 0

= 0

Let Cij be the cofactors of the elements aij in A.

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

### 6x + 15y = 13

Solution:

The given system of equations can be expressed as follows:

AX = B

Here,

A = , X =  and B =

Now,

|A| =

= 30 – 30

= 0

Let Cij be the cofactors of the elements aij in A.

C11 = 15, C12 = -6, C21 = -5 and C22 = 2

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### 6x + 9y = 10

Solution:

The given system of equations can be expressed as follows:

AX = B

Here,

A = , X =  and B =

|A| =

= 18 – 18

= 0

Let Cij be the cofactors of the elements aij in A.

C11 = 9, C12 = -6, C21 = -3 and C22 = 2

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### 6x âˆ’ 3y = 5

Solution:

The given system of equations can be expressed as,

AX = B

Here,

A = , X =  and B =

|A| =

= 12 – 12

= 0

Let Cij be the cofactors of the elements aij in A.

C11 = -3, C12 = -6, C21 = 2 and C22 = 4

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### 2x + 2y + 8z = âˆ’1

Solution:

The given system of equations can be written as,

AX = B

Here,

A = , X =  and B =

|A| =

= -144 + 180 – 36

= 0

Let Cij be the cofactors of the elements aij in A.

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### 3x âˆ’ 5y = 3

Solution:

The given system of equations can be written as,

AX = B

Here,

A = , X =  and B =

|A| =

= -15 + 3 + 12

= 0

Let Cij be the cofactors of the elements aij in A.

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### âˆ’2x + y + z = 4

Solution:

The given system of equations can be written as,

AX = B

Here,

A = , X =  and B =

|A| =

= – 3 – 3 + 6

= 0

Let Cij be the cofactors of the elements aij in A.

â‰  0

Therefore, the given system of equations is inconsistent.

Hence proved.

### Question 5. If A = and B = are two square matrices, find AB and hence solve the system of linear equations: x âˆ’ y = 3, 2x + 3y + 4z = 17, y + 2z = 7.

Solution:

Here,

A =  and B =

Now,

AB =

AB =

AB =

AB = 6I

= I

X = A-1 B

X =

Therefore x = 2, y = -1 and z = 4.

### Question 6. If A = , find Aâˆ’1 and hence solve the system of linear equations 2x âˆ’ 3y + 5z = 11, 3x + 2y âˆ’ 4z = âˆ’5, x + y + 2z = âˆ’3.

Solution:

Here,

A =

|A| =

= 0 – 6 + 5

= -1

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = – 1/- 1, y = -2/-1\ and z = -3/-1

Therefore x = 1, y = 2 and z = 3.

### Question 7. Find Aâˆ’1, if A = . Hence solve the following system of linear equations: x + 2y + 5z = 10, x âˆ’ y âˆ’ z = âˆ’2, 2x + 3y âˆ’ z = âˆ’11.

Solution:

A =

|A| =

= 4 – 2 + 25

= 27

Let Cij be the cofactors of the elements aij in A.

X = A-1 B

=> x = -27/27, y = -54/27 and z = 81/27

Therefore, x = – 1, y = -2 and z = 3.

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