System Of Linear Equations

To prove: Set of given lines are inconsistent.

Given set of lines are : -

x + 2y = 9

2x + 4y = 7

Converting the following equations in matrix form,

AX = B

R₂ – 2R₁

Again converting into equation form, we get

x + 2y = 9

0x + 0y = - 11

∴ 0 = - 11

which is not true

∴ x + 2y = 9

2x + 4y = 7 are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

2x + 3y = 5

6x + 9y = 10

Converting the following equations in matrix form,

AX = B

R₂ – 3R₁

Again converting into equation form, we get

2x + 3y = 5

0x + 0y = - 5

∴ 0 = - 5

which is not true

∴ 2x + 3y = 5

6x + 9y = 10 are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

4x – 2y = 3

6x – 3y = 5

Converting the following equations in matrix form,

AX = B

4R₂ – 6R₁

Again converting into equation form, we get

4x – 2y = 3

0x + 0y = 2

∴ 0 = 2

which is not true

∴ 4x – 2y = 3

6x – 3y = 5 are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

6x + 4y = 5

9x + 6y = 8

Converting the following equations in matrix form,

AX = B

2R₂ – 3R₁

Again converting into equation form, we get

6x + 4y = 5

0x + 0y = 3

∴ 0 = 3

which is not true

∴ 6x + 4y = 5

9x + 6y = 8 are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

x + y – 2z = 5;

x – 2y + z = - 2;

- 2x + y + z = 4

Converting the following equations in matrix form,

AX = B

=

R₂ – R₁

R₃ + 2R₁

=

R₃ + R₂

=

Converting back into equation form we get,

x + y – 2z = 5;

0x – 3y + 3z = - 7;

0x + 0y + 0z = 7

∴ 0 = 7

Which is not true.

∴x + y – 2z = 5;

x – 2y + z = - 2;

- 2x + y + z = 4

are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

2x – y + 3z = 1;

3x – 2y + 5z = - 4;

5x – 4y + 9z = 14

Converting the following equations in matrix form,

AX = B

=

2R₂ – 3R₁

2R₃ - 5R₁

=

R₃ - 3R₂

=

Converting back into equation form we get,

2x – y + 3z = 1;

0x – 1y + 1z = - 11;

0x + 0y + 0z = 56

∴ 0 = 56

Which is not true.

∴2x – y + 3z = 1;

3x – 2y + 5z = - 4;

5x – 4y + 9z = 14

are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

x + 2y + 4z = 12;

y + 2z = - 1;

3x + 2y + 4z = 4

Converting the following equations in matrix form,

AX = B

=

R₃ - 3R₁

=

R₃ + 4R₂

=

Converting back into equation form we get,

x + 2y + 4z = 12;

y + 2z = - 1;

0x + 0y + 0z = - 36

∴ 0 = - 36

Which is not true.

∴2x – y + 3z = 1;

3x – 2y + 5z = - 4;

5x – 4y + 9z = 14

are inconsistent.

To prove: Set of given lines are inconsistent.

Given set of lines are : -

3x – y – 2z = 2;

2y – z = - 1;

3x – 5y = 3

Converting the following equations in matrix form,

AX = B

=

R₃ - R₁

=

R₃ + 2R₂

=

Converting back into equation form we get,

3x – y – 2z = 2;

2y – z = - 1;

0x + 0y + 0z = - 1

∴ 0 = - 1

Which is not true.

∴3x – y – 2z = 2;

2y – z = - 1;

3x – 5y = 3

are inconsistent.

To find: - x , y

Given set of lines are : -

5x + 2y = 4;

7x + 3y = 5.

Converting the following equations in matrix form,

AX = B

5R₂ – 7R₁

Again converting into equation form, we get

5x + 2y = 4;

y = - 3

5x + 2× - 3 = 4

5x = 10

X = 2

∴ x = 2 , y = - 3

To find: - x , y

Given set of lines are : -

3x + 4y – 5 = 0;

x - y + 3 = 0

Converting the following equations in matrix form,

AX = B

3R₂ – R₁

Again converting into equation form, we get

3x + 4y = 5

- 7 y = - 14

Y = 2

3x + 4y = 5

3x + 4×2 = 5

3x = - 3

X = - 1

∴ x = - 1 , y = 2

To find: - x , y

Given set of lines are : -

x + 2y = 1

3x + y = 4

Converting the following equations in matrix form,

AX = B

R₂ – 3R₁

Again converting into equation form, we get

x + 2y = 1

- 5y = 1

Y =

x + 2 = 1

x + = 1

x = 1 +

X =

∴ x = , y =

To find: - x , y

Given set of lines are : -

5x + 7y + 2 = 0;

4x + 6y + 3 = 0.

Converting the following equations in matrix form,

AX = B

5R₂ – 4R₁

Again converting into equation form, we get

5x + 7y = - 2

2y = - 7

Y =

5x + 7 = - 2

5x = - 2 +

5x =

X =

∴

To find: - x , y

Given set of lines are : -

2x - 3y + 1 = 0;

x + 4y + 3 = 0

Converting the following equations in matrix form,

AX = B

2R₂ – R₁

Again converting into equation form we get

2x - 3y = - 1

11 y = - 5

Y =

2x - 3 = - 1

2x = - 1 -

X =

∴

To find: - x , y

Given set of lines are : -

4x - 3y = 3;

3x - 5y = 7

Converting the following equations in matrix form,

AX = B

4R₂ – 3R₁

Again converting into equation form, we get

4x – 3y = 3

- 11y = 19

Y =

4x – 3 × = 3

4x = 3 -

4x =

X =

∴

To find: - x , y , z

Given set of lines are : -

2x + 8y + 5z = 5;

x + y + z = - 2;

x + 2y - z = 2

Converting the following equations in matrix form,

AX = B

2R₂ – R₁

2R₃ – R₁

3R₃ – 2R₂

Again converting into equations, we get

2x + 8y + 5z = 5

- 6y - 3z = - 9

- 15z = 15

Z = - 1

- 6y - 3 × - 1 = - 9

- 6y = - 9 - 3

Y = 2

2x + 8×2 + 5× - 1 = 5

2x = 5 - 16 + 5

X = - 3

∴

To find: - x , y , z

Given set of lines are : -

x – y + z = 1;

2x + y – z = 2;

X – 2y – z = 4

Converting following equations in matrix form,

AX = B

R₂ – 2R₁

R₃ – R₁

3R₃ + R₂

Again converting into equations we get

X – y + z = 1

3y - 3z = 0

- 9z = 9

Z = - 1

Y = z

Y = - 1

X + 1 - 1 = 1

X = 1

∴

To find: - x , y , z

Given set of lines are : -

3X + 4y + 7z = 4;

2x – y + 3z = - 3;

x + 2y – 3z = 8

Converting the following equations in matrix form,

AX = B

3R₂ – 2R₁

3R₃ – R₁

11R₃ + 2R₂

Again converting into equations we get

3x + 4y + 7z = 4

- 11y - 5z = - 17

- 186z = 186

Z = - 1

- 11y + 5 = - 17

- 11y = - 22

Y = 2

3x + 4×2 + 7× - 1 = 4

3x = 4 - 8 + 7

X = 1

∴

To find: - x , y , z

Given set of lines are : -

x + 2y + z = 7;

x + 3z = 11;

2x – 3y = 1

Converting following equations in matrix form,

AX = B

R₂ – R₁

R₃ –2R₁

R₃ + R₂

Again converting into equations we get

X + 2y + z = 7

- 2y + 2z = 4

- 9y = - 9

Y = 1

- 2×1 + 2z = 4

2z = 6

Z = 3

X + 2×1 + 3 = 7

X = 7 - 2 - 3

X = 2

∴

To find: - x , y , z

Given set of lines are : -

2x - 3y + 5z = 16;

3x + 2y – 4z = - 4

x + y – 2z = - 3

Converting the following equations in matrix form,

AX = B

2R₂ – 3R₁

2R₃ –R₁

13R₃ - 5R₂

Again converting into equations, we get

2x - 3y + 5z = 16

13y - 23z = - 56

- 2z = - 6

Z = 3

13y – 23×3 = - 56

13y = - 56 + 69

Y = 1

2x - 3×1 + 5×3 = 16

2x = 16 + 3 - 15

2x = 4

X = 2

∴

To find: - x , y , z

Given set of lines are : -

x + y + z = 4;

2x – y + z = - 1;

2x + y – 3z = - 9.

Converting the following equations in matrix form,

AX = B

R₂ – 2R₁

R₃ –2R₁

3R₃ - R₂

Again converting into equations, we get

X + y + z = 4

- 3y - z = - 9

- 14z = - 42

Z = 3

- 3y - 3 = - 9

- 3y = - 6

Y = 2

X + 2 + 3 = 4

X = 4 - 5

X = - 1

∴

To find: - x , y , z

Given set of lines are : -

2x - 3y + 5z = 11;

3x + 2y - 4z = - 5;

x + y – 2z = - 3.

Converting the following equations in matrix form,

AX = B

2R₂ – 3R₁

2R₃ –R₁

13R₃ - 5R₂

Again converting into equations we get

2x – 3y + 5z = 11

13y - 23 z = - 43

- 2z = - 6

Z = 3

13y - 23×3 = - 43

13y = - 43 + 69

13y = 26

Y = 2

2x – 3×2 + 5×3 = 11

2x = 11 + 6 - 15

X = 1

∴

To find: - x , y , z

Given set of lines are : -

x + y + z = 1;

x – 2y + 3z = 2;

5x – 3y + z = 3.

Converting the following equations in matrix form,

AX = B

R₂ – R₁

R₃ – 5R₁

R₃ + 2R₂

Again converting into equations we get

X + y + z = 1

- 3y + 2z = 1

- 14 y = 0

Y = 0

- 3 × 0 + 2z = 1

Z =

X + 0 + = 1

X =

∴

To find: - x , y , z

Given set of lines are : -

x + y + z = 6;

x + 2z = 7;

3x + y + z = 12

Converting following equations in matrix form,

AX = B

R₂ – R₁

R₃ – 3R₁

R₃ + 2R₂

Again converting into equations we get

X + y + z = 6

- y + z = 1

- 4 y = - 4

Y = 1

- 1 + z = 1

Z = 2

X + 1 + 2 = 6

X = 6 - 3

X = 3

∴

To find: - x , y , z

Given set of lines are : -

2x + 3y + 3z = 5;

x – 2y + z = - 4;

3x – y – 2z = 3

Converting the following equations in matrix form,

AX = B

2R₂ – R₁

2R₃ – 3R₁

R₃ - 13R₂

Again converting into equations we get

2X + 3y + 3z = 5

- 7y - z = - 13

80 y = 160

Y = 2

- 7×2 - z = - 13

Z = - 1

2x + 3×2 + 3× - 1 = 5

2x = 5 - 6 + 3

X = 1

∴

To find: - x , y , z

Given set of lines are : -

4x - 5y - 11z = 12

X – 3y + z = 1;

2x + 3y – 7z = 2

Converting the following equations in matrix form,

AX = B

4R₂ – R₁

2R₃ – R₁

5R₃ + R₂

Again converting into equations we get

4x – 5y - 11z = 12

- 7y + 15z = - 8

48 y = - 48

Y = - 1

7 + 15z = - 8

15z = - 15

Z = - 1

4x + 5 + 11 = 12

4x = 12 - 5 - 11

4x = - 4

X = - 1

∴

To find: - x , y , z

Given set of lines are : -

x - y + 2z = 7

3x + 4y – 5z = - 5

2x – y + 3z = 12

Converting the following equations in matrix form,

AX = B

R₂ – 3R₁

R₃ – 2R₁

7R₃ - R₂

Again converting into equations we get

x – y + 2z = 7

7y - 11 z = - 26

4z = 12

Z = 3

7y – 11 ×3 = - 26

7y = - 26 + 33

7y = 7

Y = 1

X – 1 + 2×3 = 7

X = 7 + 1 - 6

X = 2

∴

To find: - x , y , z

Given set of lines are : -

6x - 9y - 20z = - 4

4x – 15y + 10z = - 1

2x - 3y - 5z = - 1

Converting the following equations in matrix form,

AX = B

3R₂ – 2R₁

3R₃ – R₁

Again converting into equations, we get

6x - 9y - 20 z = - 4

- 27y + 70z = 5

5z = 1

Z =

- 27y + 70 = 5

- 27y = 5 – 14

- 27y = - 9

Y =

6x – 9× - 20 × = - 4

6x = - 4 + 3 + 4

X =

∴

To find: - x , y , z

Given set of lines are : -

3x - 4y + 2z = - 1

2x + 3y + 5z = 7;

x + z = 2

Converting the following equations in matrix form,

AX = B

3R₂ – 2R₁

3R₃ – R₁

11R₃ - R₂

Again converting into equations, we get

3x - 4y + 2z = - 1

17y + 11z = 23

27 y = 54

Y = 2

17×2 + 11z = 23

11z = 23 – 34

Z = - 1

3x – 4 × 2 + 2× - 1 = - 1

3x = - 1 + 8 + 2

3x = 9

X = 3

∴

To find: - x , y , z

Given set of lines are : -

x + y - z = 1

3x + y - 2z = 3

x - y - z = - 1

Converting the following equations in matrix form,

AX = B

R₂ – 3R₁

R₃ – R₁

Again converting into equations we get

X + y – z = 1

- 2y + z = 0

- 2y = - 2

Y = 1

- 2 + z = 0

Z = 2

X + 1 - 2 = 1

X = 2

∴

To find: - x , y , z

Given set of lines are : -

2x + y - z = 1

x - y + z = 2

3x + y - 2z = - 1

Converting the following equations in matrix form,

AX = B

2R₂ – R₁

2R₃ – 3R₁

3R₃ - R₂

Again converting into equations we get

2x + y – z = 1

- 3y + 3z = 3

- 6z = - 18

Z = 3

- 3y + 3×3 = 3

- 3y = 3 - 9

- 3y = - 6

Y = 2

2x + 2 - 3 = 1

2x = 1 + 1

X = 1

∴ x = 1 , y = 2 , z = 3

To find: - x , y , z

Given set of lines are : -

x + 2y + z = 4

- x + y + z = 0

x - 3y + z = 4

Converting the following equations in matrix form,

AX = B

R₂ + R₁

R₃ – R₁

Again converting into equations we get

X + 2y + z = 4

3y + 2z = 4

- 5y = 0

Y = 0

0 + 2z = 4

Z = 2

X + 0 + 2 = 4

X = 2

∴ x = 2 , y = 0 , z = 2

To find: - x , y , z

Given set of lines are : -

x - y - 2z = 3

x + y = 1

x + z = - 6

Converting the following equations in matrix form,

AX = B

R₂ – R₁

R₃ – R₁

2R₃ - R₂

Again converting into equations we get

X + y - 2z = 3

2y + 2z = - 2

4z = - 16

Z = - 4

2y – 8 = - 2

2y = - 2 + 8

2y = 6

Y = 3

X - 3 + 8 = 3

X = - 2

∴ x = - 2 , y = 3 , z = - 4

To find: - x , y , z

Given set of lines are : -

5x - y = - 7

2x + 3z = 1

3y - z = 5

Converting the following equations in matrix form,

AX = B

5R₂ – 2R₁

2R₃ - 3R₂

Again converting into equations we get

5x – y = - 7

2y + 15z = 19

- 47z = - 47

Z = 1

2y + 15 = 19

2y = 19 - 15

Y = 2

5x – 2 = - 7

5x = - 5

X = - 1

∴ x = - 1 , y = 2 , z = 1

To find: - x , y , z

Given set of lines are : -

x - 2y + z = 0

y - z = 2

2x - 3z = 10

Converting the following equations in matrix form,

AX = B

R₃ – 2R₁

R₃ - 4R₂

Again converting into equations we get

X - 2y + z = 0

Y – z = 2

- z = 2

Z = - 2

Y + 2 = 2

Y = 0

X + 0 - 2 = 0

X = 2

∴ x = 2 , y = 0 , z = - 2

To find: - x , y , z

Given set of lines are : -

x - y = 3

2x + 3y + 4z = 17

y + 2z = 7

Converting the following equations in matrix form,

AX = B

R₂ – 2R₁

2R₃ - R₂

Again converting into equations we get

X – y = 3

5y + 4z = 11

- 3y = 3

Y = - 1

5× - 1 + 4z = 11

4z = 16

Z = 4

X + 1 = 3

X = 2

∴ x = 2 , y = - 1 , z = 4

To find: - x , y , z

Given set of lines are : -

4x + 3y + 2z = 60

x + 2y + 3z = 45

6x + 2y + 3z = 70

Converting the following equations in matrix form,

AX = B

4R₂ – R₁

2R₃ – 3R₁

Again converting into equations, we get

4x + 3y + 2z = 60

5y + 10z = 120

- 5y = - 40

Y = 8

5×8 + 10z = 120

10z = 120 - 40

10z = 80

Z = 8

4x + 3×8 + 2×8 = 60

4x = 60 – 24 - 16

4x = 20

X = 5

∴ x = 5 , y = 8 , z = 8

Given,

A =

A ^{- 1} =

The determinant of matrix A is

|A| =

= 2( 2 × - 2 – ( - 4)×1) + 3(3× - 2 – ( - 4)×1) + 5(3×1 – 2×1)

= 2( - 4 + 4 ) + 3( - 6 + 4 ) + 5( 3 – 2 )

= 2(0) + 3( - 2) + 5(1)

= - 6 + 5

= - 1

|A| ≠ 0

∴ A ^{- 1} is possible.

A^T =

Adj(A) =

A ^{- 1} =

A ^{- 1} =

A ^{- 1} =

Given set of lines are : -

2x – 3y + 5z = 11

3x + 2y – 4z = - 5

x + y – 2z = - 3

Converting following equations in matrix form,

AX = B

Where A = , X = , B =

Pre - multiplying by A ^{- 1}

A ^{- 1}AX = A ^{- 1}B

IX = A ^{- 1}B

X = A ^{- 1}B

=

=

=

∴ x = 1 , y = 2 , z = 3

Given,

A =

A ^{- 1} =

The determinant of matrix A is

|A| =

= 2( - 2× - 5 - ( - 1)×3) – (1× - 5 - ( - 1)×0) + (1×3 – ( - 2)×0)

= 2(10 + 3) – ( - 5) + (3)

= 26 + 5 + 3

= 34

|A| ≠ 0

∴ A ^{- 1} is possible.

A^T =

Adj(A) =

A ^{- 1} =

A ^{- 1} =

Given set of lines are : -

2x + y + z = 1

X – 2y – z =

3y – 5z = 9

Converting the following equations in matrix form,

AX = B

Where A = , X = , B =

Pre - multiplying by A ^{- 1}

A ^{- 1}AX = A ^{- 1}B

IX = A ^{- 1}B

X = A ^{- 1}B

=

=

= =

∴ x = 1 , y = , z = -

Given,

A = , B =

AB =

AB =

AB =

AB =

AB = 11

AB = 11I

Pre - multiplying by A ^{- 1}

A ^{- 1}AB = 11 A ^{- 1}I

IB = 11 A ^{- 1}

B = 11 A ^{- 1}

A ^{- 1} = B

Given set of lines are : -

x – 2y = 10

x + y + 3z = 8

- 2y + z = 7

Converting following equations in matrix form,

AX = C

Where A = ,X = , C =

Pre - multiplying by A ^{- 1}

A ^{- 1}AX = A ^{- 1}C

IX = A ^{- 1}C

X = A ^{- 1}C

X = BC

=

=

=

∴ x = 4 , y = - 3 , z = 1

To find: - x , y , z

Given set of lines are : -

= 10

= 10,

= 13

Converting the following equations in matrix form,

AX = B

2R₂ – R₁

2R₃ – 3R₁

R₃ - 5R₂

Again converting into equations we get

Z =

9 – 15 = 4

X =

∴ x = , y = , z =

To find: - x , y , z

Given set of lines are : -

= 4

= 0

= 2

Converting following equations in matrix form,

AX = B

R₂ – 2R₁

R₃ – R₁

Again converting into equations we get

Z = 1

- 1 - 1

X =

∴ x = , y = , z =

Let the three numbers be x, y and z.

According to the question,

X + y + z = 2

X + 2y + z = 1

5x + y + z = 6

Converting the following equations in matrix form,

AX = B

R₂ – R₁

R₃ – R₁

Converting back into the equations we get

X + y + z = 2

Y = - 1

4x = 5

X =

– 1 + z = 2

Z = 2 - + 1

Z =

∴ The numbers are .

Let the price of 1kg potato, wheat and rice be x, y and z respectively.

According to the question,

4x + 3y + 2z = 60

X + 2y + 3z = 45

6x + 2y + 3z = 70

Converting into matrix form

AX = B

4R₂ – R₁

2R₃ – 3R₁

Converting back into the equations we get

4x + 3y + 2z = 60

5y + 10z = 120

- 5y = - 40

Y = 8

5×8 + 10z = 120

10z = 120 – 40

Z = 8

4x + 3×8 + 2×8 = 60

4x = 60 - 24 - 16

4x = 20

X = 5

∴ The cost of 1 kg potatoes, wheat and rice is Rs.5, Rs.8 and Rs. 8 respectively.

Let these investments be ₹x, ₹y and ₹z, respectively.

Then, x + y + z = 5000

= 358

6x + 7y + 8z = 35800

And,

6x + 7y - 8z = 7000.

Representing in the matrix form,

AX = B

R₃ – R₂

R₂ – 6R₁

Converting back into the equations we get

X + y + z = 5000

Y + 2z = 5800

- 16z = - 28800

Z = 1800

Y + 2×1800 = 5800

Y = 5800 - 3600

Y = 2200
x + 2200 + 1800 = 5000

X = 5000 – 4000

X = 1000

Amount of 1000 , 2200 , 1800 were invested in the investments of 6% , 7%, 8% respectively.

Let the amount x, y and z be considered for sincerity, truthfulness and helpfulness.

According to the questions,

3x + 2y + z = 1600

4x + y + 3z = 2300

X + y + z = 900

Converting into the matrix form

AX = B

R₁ – 3R₃

R₂ – 4R₃

2R₂ - R₁

Converting back into the equations we get

- y - 2z = - 1100

- 5y = - 1500

X + y + z = 900

Y = 300

- 300 - 2z = - 1100

- 2z = - 800

Z = 400

X + 300 + 400 = 900

X = 900 - 700

X = 200

₹ 200 for sincerity, ₹ 300 for truthfulness and ₹ 400 for helpfulness. One more value may be like honesty, kindness, etc.

(A+B) =

(A-B) =

1+2 2A =

2A =

Dividing the matrix by 2

A =

1-2 2B =

2B =

Dividing the matrix by 2

B =

A x B=

=

=

C

+ 2A =

2A =

2A =

Dividing the matrix by 2

A =

4A + 3X = 5B

4 + 3X = 5

3X = 5-4

3X = -

3X =

Dividing by 3

X =

B

(A-2B) =

Multiplying equation by 2

2A-4B = ------------- (i)

2A-3B = ---------- (ii)

(ii)-(i)

B = -

=

(2A – B) =

_{Multiplying by 2}

4A – 2B = () ------- (i)

2B + A = () -------- (ii)

(i)+ (ii)

5A = () + ()

= ()

Dividing each element of the matrix by 5

A = ()

2

To solve this problem we will use the comparison that is we will use that all the elements of L.H.S are equal to R.H.S .

= +

=

Comparing with R.H.S

8+y = 0

y= -8

2x+1 = 5

2x = 4

x=2

A

By comparing L.H.S and R.H.S

x - y = -1 -------- i

2x – y = 0 -------- ii

2x+ z = 5 --------- iii

3z + w =13 ------- iv

Using i in equation ii

x = -1 + y

ii becomes, -2 + 2y –y = 0

y = 2

x = 1

Putting x in iii

2 + z =5

z = 3

Putting z in iv

9 + w = 13

w = 4

C

=

=

Comparing with R.H.S

x + 2y = 3 ------- (i)

2x + 3y = 5 -------- (ii)

(i) x 2 – (ii)

2x + 4y – 2x + 3y = 6 - 5

y = 1

Putting y in (i)

x + 2(1) = 3

x= 1

When a given matrix is singular then the given matrix determinant is 0.

= 0

Given, A =

= 0

_{4(3-2x) – 2(x+1) = 0}

_{12 – 8x -2x -2 =0}

_{10 -10x= 0}

_{10x = 0}

_{x= 1}

Given, A_α =

A_α²₌

₌

₌

₌

L.H.S: A + A^’ =

=

=

This will be equal to

When 2cosa = 1

cosa =

a =

When a given matrix is singular then the given matrix determinant is 0.

= 0

Given,

A=

= 0

1(-4k + 6) –k(-12 + 4) +3 (9 -2k)= 0

-4k + 6 +12k -4k + 27 -6k = 0

-2k +33 = 0

k = .

To find adj A we will first find the cofactor matrix

C₁₁ = d C₁₂ = -c

C₂₁= -b C₂₂ = a

Cofactor matrix A =

Adj A =’

=

We know that A x A^-1 = I

=

=

To satisfy the above condition 2x = 1

x =

Since A and B are square matrices of same order.

(A+B)(A-B) = A² –AB +BA – B

Since A and B are square matrices of same order.

(A + B)² = (A+ B)(A + B)

= A² + AB + BA+ B²

Since A and B are square matrices of same order.

(A - B)² = (A- B)(A - B)

= A² - AB - BA+ B²

Given A and B are symmetric matrices

A’ = A --- 1

B’ = B ---- 2

Now (AB – BA)’ = (AB)’ – (BA)’

=B’A’ – A’B’

[’ = B’A’ ]

= BA – AB [Using 1 and 2]

(AB – BA)’ = - (AB - BA)

AB-BA is a skew symmetric matrix.

A = B^-1

B=A^-1

We know that

AA^-1= I

(Given B=A^-1)

AB= I ------ 1

We know that

BB^-1= I

(Given A=B^-1)

BA= I ------ 2

From 1 and 2

AB= BA = I

We know that (AB)^-1 = adj(AB)/

adj (AB)= (AB)^-1

We also know that (AB)^-1 = B^-1. A^-1

Putting them in 1

Adj (AB) = B^-1. A^-1.

= (A^-1.) (B^-1)

= adj(A) adj(B)

Since, adj (A)= (A)^-1

adj (B)= (B)^-1

The property states that

adj(adj A) = ^n-2 . A

Here n=2

adj(adj A) = ^3-2 . A

= 4A

The property states that │adj A│= │A│^n-1

Here n= 3 and │A│=5

│adj A│= │5│^3-1

= │5│²

= 25.

If the two matrices A and B are of same order it is not necessary that in every situation AB= BA

AB= BA = I only when A = B^-1

B=A^-1

Other time ABBA

The matrix on the R.H.S of the given matrix is of order 2 x 2 and the one given on left side is 2 x 2 . Therefore A has to be a 2 x 2 matrix.

Let A =

=

=

3a+b = 4 ----- 1

2a-b = 1 ------ 2

3c+d= 2 ------ 3

2c-d=3 ------- 4

Using 1 and 2

a=1

b=1

Using 3 and 4

c=1

d = -1

So A becomes

B

We know that AA^-1 =

Taking determinant both sides

│AA^-1│ = │I│

│A││A^-1│ = │I│ (│AB│=│A││B│)

│A││A^-1│ = 1 (│I│ = 1)

│A^-1│=

(AB)(AB)^-1 = I

A^-1(AB)(AB)^-1 = IA^-1

(A^-1A)B (AB)^-1=A^-1

IB(AB)^-1 = A^-1

B(AB)^-1 = A^-1

B^-1B(AB)^-1 = B^-1 A^-1

I (AB)^-1 = B^-1A^-1

(AB)^-1 = B^-1A^-1

s AB is a 0 matrix its determinant has to be 0.

So │AB│=│A││B│=0

So │A│=│B│=0

² – A + 2I = 0

Multiplying by A^-1

A^-1A² – A^-1A + 2I A^-1 = 0

A- I + 2 A^-1 = 0

A^-1=

=

│A│=0

1(2 x 1 – 5 x1) - (1 x 1 – 5 x 2) + 2 ( 1x1 – 2x 2) = 0

-3+9 λ -6 = 0

9 λ = 9

λ = 1

A=

│A│= cos² - (-sin²)

= cos² + (sin²)

= 1 ---------- (I)

We know that A^-1= adj A

= adj A [From I]

Matrix A is said to be nilpotent since there exist a positive integer k=1 such that Ak is zero matrix.

Here the diagonal value is 2+3-3= 1

So the given matrix is idempotent.

A(adjA)= A(│A│ x A^-1)

Since determinant of singular matrix is always 0

A(adjA)= 0

So, it is a null matrix.

(adjA) =

= 8

= │A│I

│A│= 8.

A =

│A│= -2-3 = -5

We know that │A^-1│=

=

² + xI = yA

+ x = y

+x = y

8+x = y

Comparing L.H.S and R.H.S

x=8 y=8

If A and B anticommute then AB= -BA

To find adj A we will first find the cofactor matrix

C₁₁ = 3 C₁₂ = -1

C₂₁= -5 C₂₂ = 2

Cofactor matrix A =

Adj A =’

=

B=I

B = A^-1 I ---------1

A^-1= adj A --------- 2

= 3 x 2 – (-4) x (-1)

2

C₁₁ = 2 C₁₂ = 1

C₂₁= 4 C₂₂ = 3

Cofactor matrix A =

Adj A =’

=

Putting in 2

A^-1=

=

Putting in 1

B = A^-1 I

= A^-1

=

(AB)(AB)^-1 = I

A^-1(AB)(AB)^-1 = IA^-1

(A^-1A)B (AB)^-1=A^-1

IB(AB)^-1 = A^-1

B(AB)^-1 = A^-1

B^-1B(AB)^-1 = B^-1 A^-1

I (AB)^-1 = B^-1A^-1

(AB)^-1 = B^-1A^-1

^-1= adj A --------- 1

= 3 x 2 – (1) x (-1)

7

C₁₁ = 3 C₁₂ = -1

C₂₁= 1 C₂₂ = 2

Cofactor matrix A =

Adj A =’

=

Putting in 1

A^-1=

=

^-1= adj A

adj A = x A^-1

= 3 x

=

y property of inverse

(A^-1)^-1 = A

(A^-1)^-1 =^-1

A =^-1 ------------ 1

^-1 = 3 x 6 – 4 x 5

-2

C₁₁ = 6 C₁₂ = -5

C₂₁= -4 C₂₂ = 3

Cofactor matrix A =

Adj A =

^-1 =

=

Putting in 1

A =

ƒ(A) = 2A² – 4A + 5

A² =

=

ƒ(A) = 2A² – 4A + 5I

= 2 - 4 +5

= - +

=

A² =

=

A² – 4A = -4

= -

=

= 5

= 5I

(adj A) = │A│I

= 6

=

y the property of inverse

(AB)^-1 = B^-1A^-1

(KA)^-1= A^-1K^-1

= A^-1

= 3 x (0 - 2) – 4 x (2-4)+1 x (-1)

= -6+8-1

= 1

C₁₁ = -2 C₁₂ = 2 C₁₃= -1

C₂₁= -9 C₂₂ = 8 C₂₃= -5

C₃₁= -8 C₃₂=7 C₃₃ =-4

Cofactor (A)= [ ]

Adj A = [ ]’

=[ ]

A^-1= adj A

= [ ]

Let X = A+A’

X’ = (A+A’)’

= A’ + (A’)’

=A + A’

= X

Therefore (A+A’) is symmetric matrix.

Let X = A-A’

X’ = (A-A’)’

= A’ - (A’)’

=A’ – A

= -(A – A’)

= -X

Therefore (A-A’) is skew symmetric matrix.

Since the matrix is of order 3 so 3 will be taken common from each row or column.

So, k= 27

Tagging

=

= -8

Since -8 could be taken common from each row or column. Hence C is a scalar matrix.

= B=

A+B =

(A+B)² =

=

=

=

A²=

=

B²=

=

(A + B)² = (A² + B²)

= +

=

By comparison,

a-1 = 0

a=1

b=4