Chapter 6 Determinants Ex 6.1

1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Solution:

(i) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

From the given matrix we have,

M₁₁ = –1

M₂₁ = 20

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –1

= –1

C₂₁ = (–1)²⁺¹ × M₂₁

= 20 × –1

= –20

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= 5× (–1) + 0 × (–20)

= –5

(ii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given

From the above matrix we have

M₁₁ = 3

M₂₁ = 4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 3

= 3

C₂₁ = (–1)²⁺¹ × 4

= –1 × 4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= –1× 3 + 2 × (–4)

= –11

(iii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = –3 × 2 – (–1) × 2

M₃₁ = –4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × –12

= –12

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –16

= 16

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (–12) + 4 × 16 + 3× (–4)

= –12 + 64 –12

= 40

(iv) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = a × c a – b × bc

M₃₁ = a²c – b²c

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (ab² – ac²)

= ab² – ac²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (a²b – c²b)

= c²b – a²b

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (a²c – b²c)

= a²c – b²c

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c)

= ab² – ac² + c²b – a²b + a²c – b²c

(v) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = 2×0 – 5×6

M₃₁ = –30

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 5

= 5

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –40

= 40

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –30

= –30

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 0× 5 + 1 × 40 + 3× (–30)

= 0 + 40 – 90

= 50

(vi) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = h × f – b × g

M₃₁ = hf – bg

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (bc– f²)

= bc– f²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (hc – fg)

= fg – hc

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (hf – bg)

= hf – bg

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= a× (bc– f²) + h× (fg – hc) + g× (hf – bg)

= abc– af² + hgf – h²c +ghf – bg²

(vii) Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also, C_ij = (–1)^i+j × M_ij

Given,

M₃₁ = –1(1 × 0 – 5 × (–2)) – 0(0 × 0 – (–1) × (–2)) + 1(0 × 5 – (–1) × 1)

M₃₁ = –9

M₄₁ = –1(1×1 – (–1) × (–2)) – 0(0 × 1 – 1 × (–2)) + 1(0 × (–1) – 1 × 1)

M₄₁ = 0

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (–9)

= –9

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × 9

= –9

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –9

= –9

C₄₁ = (–1)⁴⁺¹ × M₄₁

= –1 × 0

= 0

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁

= 2 × (–9) + (–3) × –9 + 1 × (–9) + 2 × 0

= – 18 + 27 –9

= 0

2. Evaluate the following determinants:

Solution:

(i) Given

⇒ |A| = x (5x + 1) – (–7) x

|A| = 5x² + 8x

(ii) Given

⇒ |A| = cos θ × cos θ – (–sin θ) x sin θ

|A| = cos²θ + sin²θ

We know that cos²θ + sin²θ = 1

|A| = 1

(iii) Given

⇒ |A| = cos15° × cos75° + sin15° x sin75°

We know that cos (A – B) = cos A cos B + Sin A sin B

By substituting this we get, |A| = cos (75 – 15)°

|A| = cos60°

|A| = 0.5

(iv) Given

⇒ |A| = (a + ib) (a – ib) – (c + id) (–c + id)

= (a + ib) (a – ib) + (c + id) (c – id)

= a² – i² b² + c² – i² d²

We know that i² = -1

= a² – (–1) b² + c² – (–1) d²

= a² + b² + c² + d²

3. Evaluate:

Solution:

Since |AB|= |A||B|

= 2(17 × 12 – 5 × 20) – 3(13 × 12 – 5 × 15) + 7(13 × 20 – 15 × 17)

= 2 (204 – 100) – 3 (156 – 75) + 7 (260 – 255)

= 2×104 – 3×81 + 7×5

= 208 – 243 +35

= 0

Now |A|² = |A|×|A|

|A|²= 0

4. Show that

Solution:

Given

Let the given determinant as A

Using sin (A+B) = sin A × cos B + cos A × sin B

⇒ |A| = sin 10° × cos 80° + cos 10° x sin 80°

|A| = sin (10 + 80)°

|A| = sin90°

|A| = 1

Hence Proved

Solution:

Given,

= 2(1 × 1 – 4 × (–2)) – 3(7 × 1 – (–2) × (–3)) – 5(7 × 4 – 1 × (–3))

= 2(1 + 8) – 3(7 – 6) – 5(28 + 3)

= 2 × 9 – 3 × 1 – 5 × 31

= 18 – 3 – 155

= –140

Now by expanding along the second column

= 2(1 × 1 – 4 × (–2)) – 7(3 × 1 – 4 × (–5)) – 3(3 × (–2) – 1 × (–5))

= 2 (1 + 8) – 7 (3 + 20) – 3 (–6 + 5)

= 2 × 9 – 7 × 23 – 3 × (–1)

= 18 – 161 +3

= –140

Solution:

Given

⇒ |A| = 0 (0 – sinβ (–sinβ)) –sinα (–sinα × 0 – sinβ cosα) – cosα ((–sinα) (–sinβ) – 0 × cosα)

|A| = 0 + sinα sinβ cosα – cosα sinα sinβ

|A| = 0

Chapter 6 Determinants Ex 6.2

1. Evaluate the following determinant:

Solution:

(i) Given

(ii) Given

= 1(109)(12)–(119)(11)

= 1308 – 1309

= – 1

So, Δ = – 1

(iii) Given,

= a (bc – f²) – h (hc – fg) + g (hf – bg)

= abc – af² – ch² + fgh + fgh – bg²

= abc + 2fgh – af² – bg² – ch²

So, Δ = abc + 2fgh – af² – bg² – ch²

(iv) Given

= 21(24–4) = 40

So, Δ = 40

(v) Given

= 1(–7)(–36)–(–20)(–13)

= 252 – 260

= – 8

So, Δ = – 8

(vi) Given,

(vii) Given

(viii) Given,

2. Without expanding, show that the value of each of the following determinants is zero:

Solution:

(i) Given,

(ii) Given,

(iii) Given,

(iv) Given,

(v) Given,

(vi) Given,

(vii) Given,

(viii) Given,

(ix) Given,

As, C₁ = C₂, hence determinant is zero

(x) Given,

(xi) Given,

(xii) Given,

(xiii) Given,

(xiv) Given,

(xv) Given,

(xvi) Given,

(xvii) Given,

Hence proved.

Evaluate the following (3 – 9):

Solution:

Given,

= (a + b + c) (b – a) (c – a) (b – c)

So, Δ = (a + b + c) (b – a) (c – a) (b – c)

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

= a a(a+x+y)+az + 0 + 0

= a² (a + x + y + z)

So, Δ = a² (a + x + y + z)

Solution:

Prove the following identities (11 – 45):

Solution:

Given,

Solution:

Consider,

= – (a + b + c) (b–c)(a+b–2c)–(c–a)(c+a–2b)

= 3abc – a³ – b³ – c³

Therefore, L.H.S = R.H.S,

Hence the proof.

Solution:

Given,

Solution:

Consider,

,

Solution:

Consider,

L.H.S =

Now by applying, R₁→R₁ + R₂ + R₃, we get,

Solution:

Consider,

Solution:

Consider,

Solution:

Consider,

Hence, the proof.

Solution:

Given,

= – xyz(x – y) (z – y) [z² + y² + zy – x² – y² – xy]

= – xyz(x – y) (z – y) (z–x)(z+x0+y(z–x)

= – xyz(x – y) (z – y) (z – x) (x + y + z)

= R.H.S

Hence, the proof.

Solution:

Consider,

= (a² + b² + c²) (b – a) (c – a) (b+a)(–b)–(–c)(c+a)

= (a² + b² + c²) (a – b) (c – a) (b – c) (a + b + c)

= R.H.S

Hence, the proof.

Solution:

Consider,

= (2a+4)(1)–(1)(2a+6)

= – 2

= R.H.S

Hence, the proof.

Solution:

Consider,

= – (a² + b² + c²) (a – b) (c – a) (–(b+a))(–b)–(c)(c+a)

= (a – b) (b – c) (c – a) (a + b + c) (a² + b² + c²)

= R.H.S

Hence, the proof.

Solution:

Consider,

= R.H.S

Hence, the proof.

Solution:

Consider,

Solution:

Consider,

Solution:

Expanding the determinant along R₁, we have

Δ = 1(1)(7)–(3)(2) – 0 + 0

∴ Δ = 7 – 6 = 1

Thus,

Hence the proof.

Chapter 6 Determinants Ex 6.3

1. Find the area of the triangle with vertices at the points:

(i) (3, 8), (-4, 2) and (5, -1)

(ii) (2, 7), (1, 1) and (10, 8)

(iii) (-1, -8), (-2, -3) and (3, 2)

(iv) (0, 0), (6, 0) and (4, 3)

Solution:

(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.

We know that, if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

As we know area cannot be negative. Therefore, 15 square unit is the area

Thus area of triangle is 15 square units

(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.

We know that if vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by:

2. Using the determinants show that the following points are collinear:

(i) (5, 5), (-5, 1) and (10, 7)

(ii) (1, -1), (2, 1) and (10, 8)

(iii) (3, -2), (8, 8) and (5, 2)

(iv) (2, 3), (-1, -2) and (5, 8)

Solution:

(i) Given (5, 5), (-5, 1) and (10, 7)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by

(ii) Given (1, -1), (2, 1) and (10, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

(iii) Given (3, -2), (8, 8) and (5, 2)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

Now, by substituting given value in above formula

Since, Area of triangle is zero

Hence, points are collinear.

(iv) Given (2, 3), (-1, -2) and (5, 8)

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab

Solution:

Given (a, 0), (0, b) and (1, 1) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒

⇒ a + b = ab

Hence Proved

4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b.

Solution:

Given (a, b), (a’, b’) and (a – a’, b – b) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ a b’ = a’ b

Hence, the proof.

5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear.

Solution:

Given (1, -5), (-4, 5) and (λ, 7) are collinear

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ – 50 – 10λ = 0

⇒ λ = – 5

6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).

Solution:

Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the area of the triangle is given by,

⇒ x(–10)–4(–3)+1(8–30) = ± 70

⇒ –10x+12+38 = ± 70

⇒ ±70 = – 10x + 50

Taking positive sign, we get

⇒ + 70 = – 10x + 50

⇒ 10x = – 20

⇒ x = – 2

Taking –negative sign, we get

⇒ – 70 = – 10x + 50

⇒ 10x = 120

⇒ x = 12

Thus x = – 2, 12

Chapter 6 Determinants Ex 6.4

Solve the following system of linear equations by Cramer’s rule:

1. x – 2y = 4

-3x + 5y = -7

Solution:

Given x – 2y = 4

-3x + 5y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 5(1) – (– 3) (– 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 5(4) – (– 7) (– 2)

⇒ D₁ = 20 – 14

⇒ D₁ = 6

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(– 7) – (– 3) (4)

⇒ D₂ = – 7 + 12

⇒ D₂ = 5

Thus by Cramer’s Rule, we have

2. 2x – y = 1

7x – 2y = -7

Solution:

Given 2x – y = 1 and

7x – 2y = -7

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(– 2) – (– 7) (– 1)

⇒ D₁ = – 2 – 7

⇒ D₁ = – 9

And

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(– 7) – (7) (1)

⇒ D₂ = – 14 – 7

⇒ D₂ = – 21

Thus by Cramer’s Rule, we have

3. 2x – y = 17

3x + 5y = 6

Solution:

Given 2x – y = 17 and

3x + 5y = 6

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₁ = 17(5) – (6) (– 1)

⇒ D₁ = 85 + 6

⇒ D₁ = 91

Solving determinant, expanding along 1^st row

⇒ D₂ = 2(6) – (17) (3)

⇒ D₂ = 12 – 51

⇒ D₂ = – 39

Thus by Cramer’s Rule, we have

4. 3x + y = 19

3x – y = 23

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(– 1) – (3) (1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 19(– 1) – (23) (1)

⇒ D₁ = – 19 – 23

⇒ D₁ = – 42

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(23) – (19) (3)

⇒ D₂ = 69 – 57

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

5. 2x – y = -2

3x + 4y = 3

Solution:

Given 2x – y = -2 and

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (– 2) (3)

⇒ D₂ = 6 + 6

⇒ D₂ = 12

Thus by Cramer’s Rule, we have

6. 3x + ay = 4

2x + ay = 2, a ≠ 0

Solution:

Given 3x + ay = 4 and

2x + ay = 2, a ≠ 0

Let there be a system of n simultaneous linear equations and with n unknown given by

3x + ay = 4

2x + ay = 2, a≠0

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 3(a) – (2) (a)

⇒ D = 3a – 2a

⇒ D = a

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 4(a) – (2) (a)

⇒ D = 4a – 2a

⇒ D = 2a

Solving determinant, expanding along 1^st row

⇒ D₂ = 3(2) – (2) (4)

⇒ D = 6 – 8

⇒ D = – 2

Thus by Cramer’s Rule, we have

7. 2x + 3y = 10

x + 6y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 2 (6) – (3) (1)

⇒ D = 12 – 3

⇒ D = 9

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10 (6) – (3) (4)

⇒ D = 60 – 12

⇒ D = 48

Solving determinant, expanding along 1^st row

⇒ D₂ = 2 (4) – (10) (1)

⇒ D₂ = 8 – 10

⇒ D₂ = – 2

Thus by Cramer’s Rule, we have

8. 5x + 7y = -2

4x + 6y = -3

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x + 7y = – 2

4x + 6y = – 3

So by comparing with the theorem, let’s find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 5(6) – (7) (4)

⇒ D = 30 – 28

⇒ D = 2

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = – 2(6) – (7) (– 3)

⇒ D₁ = – 12 + 21

⇒ D₁ = 9

Solving determinant, expanding along 1^st row

⇒ D₂ = – 3(5) – (– 2) (4)

⇒ D₂ = – 15 + 8

⇒ D₂ = – 7

Thus by Cramer’s Rule, we have

9. 9x + 5y = 10

3y – 2x = 8

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 3(9) – (5) (– 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 10(3) – (8) (5)

⇒ D₁ = 30 – 40

⇒ D₁ = – 10

Solving determinant, expanding along 1^st row

⇒ D₂ = 9(8) – (10) (– 2)

⇒ D₂ = 72 + 20

⇒ D₂ = 92

Thus by Cramer’s Rule, we have

10. x + 2y = 1

3x + y = 4

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Solving determinant, expanding along 1^st row

⇒ D = 1(1) – (3) (2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 1(1) – (2) (4)

⇒ D₁ = 1 – 8

⇒ D₁ = – 7

Solving determinant, expanding along 1^st row

⇒ D₂ = 1(4) – (1) (3)

⇒ D₂ = 4 – 3

⇒ D₂ = 1

Thus by Cramer’s Rule, we have

Solve the following system of linear equations by Cramer’s rule:

11. 3x + y + z = 2

2x – 4y + 3z = -1

4x + y – 3z = -11

Solution:

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let’s find D, D₁, D₂ and D₃

Solving determinant, expanding along 1^st row

⇒ D = 3(–4)(–3)–(3)(1) – 1(2)(–3)–12 + 12–4(–4)

⇒ D = 312–3 – –6–12 + 2+16

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 2(–4)(–3)–(3)(1) – 1(–1)(–3)–(–11)(3) + 1(–1)–(–4)(–11)

⇒ D₁ = 212–3 – 13+33 + 1–1–44

⇒ D₁ = 29 – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 33+33 – 2–6–12 + 1–22+4

⇒ D₂ = 336 – 2(– 18) – 18

⇒ D₂ = 126

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 344+1 – 1–22+4 + 22+16

⇒ D₃ = 345 – 1(– 18) + 2(18)

⇒ D₃ = 135 + 18 + 36

⇒ D₃ = 189

Thus by Cramer’s Rule, we have

12. x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Solution:

Given,

x – 4y – z = 11

2x – 5y + 2z = 39

-3x + 2y + z = 1

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

x – 4y – z = 11

2x – 5y + 2z = 39

– 3x + 2y + z = 1

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st row

⇒ D = 1(–5)(1)–(2)(2) + 4(2)(1)+6 – 14+5(–3)

⇒ D = 1–5–4 + 48 – –11

⇒ D = – 9 + 32 + 11

⇒ D = 34

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 11(–5)(1)–(2)(2) + 4(39)(1)–(2)(1) – 12(39)–(–5)(1)

⇒ D₁ = 11–5–4 + 439–2 – 178+5

⇒ D₁ = 11–9 + 4(37) – 83

⇒ D₁ = – 99 – 148 – 45

⇒ D₁ = – 34

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 139–2 – 112+6 – 12+117

⇒ D₂ = 137 – 11(8) – 119

⇒ D₂ = – 170

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 1–5–(39)(2) – (– 4) 2–(39)(–3) + 114–(–5)(–3)

⇒ D₃ = 1 –5–78 + 4 (2 + 117) + 11 (4 – 15)

⇒ D₃ = – 83 + 4(119) + 11(– 11)

⇒ D₃ = 272

Thus by Cramer’s Rule, we have

13. 6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Solution:

Given

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

So by comparing with theorem, now we have to find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 6(4)(3)–(1)(–2) – 1(4)(1)+4 – 31–3(2)

⇒ D = 612+2 – 8 – 3–5

⇒ D = 84 – 8 + 15

⇒ D = 91

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5(4)(3)–(–2)(1) – 1(5)(4)–(–2)(8) – 3(5)–(3)(8)

⇒ D₁ = 512+2 – 120+16 – 35–24

⇒ D₁ = 514 – 36 – 3(– 19)

⇒ D₁ = 70 – 36 + 57

⇒ D₁ = 91

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 620+16 – 54–2(–2) + (– 3)8–10

⇒ D₂ = 636 – 5(8) + (– 3) (– 2)

⇒ D₂ = 182

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 624–5 – 18–10 + 51–6

⇒ D₃ = 619 – 1(– 2) + 5(– 5)

⇒ D₃ = 114 + 2 – 25

⇒ D₃ = 91

Thus by Cramer’s Rule, we have

14. x + y = 5

y + z = 3

x + z = 4

Solution:

Given x + y = 5

y + z = 3

x + z = 4

Let there be a system of n simultaneous linear equations and with n unknown given by

Let D_j be the determinant obtained from D after replacing the j^th column by

Now, here we have

x + y = 5

y + z = 3

x + z = 4

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 11 – 1–1 + 0–1

⇒ D = 1 + 1 + 0

⇒ D = 2

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 51 – 1(3)(1)–(4)(1) + 00–(4)(1)

⇒ D₁ = 5 – 13–4 + 0–4

⇒ D₁ = 5 – 1–1 + 0

⇒ D₁ = 5 + 1 + 0

⇒ D₁ = 6

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 13–4 – 5–1 + 00–3

⇒ D₂ = 1–1 + 5 + 0

⇒ D₂ = 4

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 14–0 – 10–3 + 50–1

⇒ D₃ = 14 – 1(– 3) + 5(– 1)

⇒ D₃ = 4 + 3 – 5

⇒ D₃ = 2

Thus by Cramer’s Rule, we have

15. 2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Solution:

Given

2y – 3z = 0

x + 3y = -4

3x + 4y = 3

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 00 – 2(0)(1)–0 – 31(4)–3(3)

⇒ D = 0 – 0 – 34–9

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 00 – 2(0)(–4)–0 – 34(–4)–3(3)

⇒ D₁ = 0 – 0 – 3–16–9

⇒ D₁ = 0 – 0 – 3(– 25)

⇒ D₁ = 0 – 0 + 75

⇒ D₁ = 75

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant

⇒ D₂ = 00 – 0(0)(1)–0 – 31(3)–3(–4)

⇒ D₂ = 0 – 0 + (– 3) (3 + 12)

⇒ D₂ = – 45

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 09–(–4)4 – 2(3)(1)–(–4)(3) + 01(4)–3(3)

⇒ D₃ = 025 – 2(3 + 12) + 0(4 – 9)

⇒ D₃ = 0 – 30 + 0

⇒ D₃ = – 30

Thus by Cramer’s Rule, we have

16. 5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Solution:

Given

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Let there be a system of n simultaneous linear equations and with n unknown given by

Now, here we have

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

So by comparing with theorem, now we have to find D, D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 5(–8)(–6)–(–1)(2) – 7(–6)(6)–3(–1) + 12(6)–3(–8)

⇒ D = 548+2 – 7–36+3 + 112+24

⇒ D = 250 – 231 + 36

⇒ D = 55

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 11(–8)(–6)–(2)(–1) – (– 7) (15)(–6)–(–1)(7) + 1(15)2–(7)(–8)

⇒ D₁ = 1148+2 + 7–90+7 + 130+56

⇒ D₁ = 1150 + 7–83 + 86

⇒ D₁ = 550 – 581 + 86

⇒ D₁ = 55

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 5(15)(–6)–(7)(–1) – 11 (6)(–6)–(–1)(3) + 1(6)7–(15)(3)

⇒ D₂ = 5–90+7 – 11–36+3 + 142–45

⇒ D₂ = 5–83 – 11(– 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₃ = 5(–8)(7)–(15)(2) – (– 7) (6)(7)–(15)(3) + 11(6)2–(–8)(3)

⇒ D₃ = 5–56–30 – (– 7) 42–45 + 1112+24

⇒ D₃ = 5–86 + 7–3 + 1136

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have

Chapter 6 Determinants Ex 6.5

Solve each of the following system of homogeneous linear equations:

1. x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Solution:

Given x + y – 2z = 0

2x + y – 3z =0

5x + 4y – 9z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

= 1(1 × (– 9) – 4 × (– 3)) – 1(2 × (– 9) – 5 × (– 3)) – 2(4 × 2 – 5 × 1)

= 1(– 9 + 12) – 1(– 18 + 15) – 2(8 – 5)

= 1 × 3 –1 × (– 3) – 2 × 3

= 3 + 3 – 6

= 0

Since D = 0, so the system of equation has infinite solution.

Now let z = k

⇒ x + y = 2k

And 2x + y = 3k

Now using the Cramer’s rule

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

x + y + z = 0

2x + 5y – 2z = 0

Solution:

Given

2x + 3y + 4z = 0

x + y + z = 0

2x + 5y – 2z = 0

Any system of equation can be written in matrix form as AX = B

Now finding the Determinant of these set of equations,

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

= 2(– 2 – 5) – 3(– 2 – 2) + 4(5 – 2)

= 1 × (– 7) – 3 × (– 4) + 4 × 3

= – 7 + 12 + 12

= 17

Since D ≠ 0, so the system of equation has infinite solution.

Therefore the system of equation has only solution as x = y = z = 0.

Chapter 5 Algebra of Matrices Ex 5.1

1. If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Solution:

If a matrix is of order m × n elements, it has m n elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

m n = 8

Then, ordered pairs m and n will be

m × n be (8 × 1),(1 × 8),(4 × 2),(2 × 4)

Now, if it has 5 elements

Possible orders are (5 × 1), (1 × 5).

Solution:

(i)

Now, Comparing with equation (1) and (2)

a₂₂ = 4 and b₂₁ = – 3

a₂₂ + b₂₁ = 4 + (– 3) = 1

(ii)

Now, Comparing with equation (1) and (2)

a₁₁ = 2, a₂₂ = 4, b₁₁ = 2, b₂₂ = 4

a₁₁ b₁₁ + a₂₂ b₂₂ = 2 × 2 + 4 × 4 = 4 + 16 = 20

3. Let A be a matrix of order 3 × 4. If R₁ denotes the first row of A and C₂ denotes its second column, then determine the orders of matrices R₁ and C₂.

Solution:

Given A be a matrix of order 3 × 4.

So, A = [a_{i j}] _3×4

R₁ = first row of A = [a₁₁, a₁₂, a₁₃, a₁₄]

So, order of matrix R₁ = 1 × 4

C₂ = second column of

Therefore order of C₂ = 3 × 1

4. Construct a 2 ×3 matrix A = [a_{j j}] whose elements a_{j j} are given by:

(i) a_{i j} = i × j

(ii) a_{i j} = 2i – j

(iii) a_{i j} = i + j

(iv) a_{i j} = (i + j)²/2

Solution:

(i) Given a_{i j} = i × j

Let A = [a_{i j}]_{2 × 3}

So, the elements in a 2 × 3 matrix are[a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃]

a₁₁ = 1 × 1 = 1

a₁₂ = 1 × 2 = 2

a₁₃ = 1 × 3 = 3

a₂₁ = 2 × 1 = 2

a₂₂ = 2 × 2 = 4

a₂₃ = 2 × 3 = 6

Substituting these values in matrix A we get,

(ii) Given a_{i j} = 2i – j

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 2 × 1 – 1 = 2 – 1 = 1

a₁₂ = 2 × 1 – 2 = 2 – 2 = 0

a₁₃ = 2 × 1 – 3 = 2 – 3 = – 1

a₂₁ = 2 × 2 – 1 = 4 – 1 = 3

a₂₂ = 2 × 2 – 2 = 4 – 2 = 2

a₂₃ = 2 × 2 – 3 = 4 – 3 = 1

Substituting these values in matrix A we get,

(iii) Given a_{i j} = i + j

Let A = [a _{i j}] _2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

Substituting these values in matrix A we get,

(iv) Given a_{i j} = (i + j)²/2

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

Substituting these values in matrix A we get,

5. Construct a 2 × 2 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) (i + j)² /2

(ii) a_{i j} = (i – j)² /2

(iii) a_{i j} = (i – 2j)² /2

(iv) a_{i j} = (2i + j)² /2

(v) a_{i j} = |2i – 3j|/2

(vi) a_{i j} = |-3i + j|/2

(vii) a_{i j} = e^2ix sin x j

Solution:

(i) Given (i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(ii) Given a_{i j} = (i – j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(iii) Given a_{i j} = (i – 2j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(iv) Given a_{i j} = (2i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(v) Given a_{i j} = |2i – 3j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(vi) Given a_{i j} = |-3i + j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

(vii) Given a_{i j} = e^2ix sin x j

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A we get,

6. Construct a 3×4 matrix A = [a_{i j}] whose elements a_{i j} are given by:
(i) a_{i j} = i + j

(ii) a_{i j} = i – j

(iii) a_{i j} = 2i

(iv) a_{i j} = j

(v) a_{i j} = ½ |-3i + j|

Solution:

(i) Given a_{i j} = i + j

Let A = [a_{i j}]_2×3

So, the elements in a 3 × 4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₁₄ = 1 + 4 = 5

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

a_{24 =} 2 + 4 = 6

a₃₁ = 3 + 1 = 4

a₃₂ = 3 + 2 = 5

a₃₃ = 3 + 3 = 6

a₃₄ = 3 + 4 = 7

Substituting these values in matrix A we get,

A =

(ii) Given a_{i j} = i – j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 – 1 = 0

a₁₂ = 1 – 2 = – 1

a₁₃ = 1 – 3 = – 2

a₁₄ = 1 – 4 = – 3

a₂₁ = 2 – 1 = 1

a₂₂ = 2 – 2 = 0

a₂₃ = 2 – 3 = – 1

a_{24 =} 2 – 4 = – 2

a₃₁ = 3 – 1 = 2

a₃₂ = 3 – 2 = 1

a₃₃ = 3 – 3 = 0

a₃₄ = 3 – 4 = – 1

Substituting these values in matrix A we get,

A =

(iii) Given a_{i j} = 2i

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 2×1 = 2

a₁₂ = 2×1 = 2

a₁₃ = 2×1 = 2

a₁₄ = 2×1 = 2

a₂₁ = 2×2 = 4

a₂₂ = 2×2 = 4

a₂₃ = 2×2 = 4

a_{24 =} 2×2 = 4

a₃₁ = 2×3 = 6

a₃₂ = 2×3 = 6

a₃₃ = 2×3 = 6

a₃₄ = 2×3 = 6

Substituting these values in matrix A we get,

A =

(iv) Given a_{i j} = j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1

a₁₂ = 2

a₁₃ = 3

a₁₄ = 4

a₂₁ = 1

a₂₂ = 2

a₂₃ = 3

a_{24 =} 4

a₃₁ = 1

a₃₂ = 2

a₃₃ = 3

a₃₄ = 4

Substituting these values in matrix A we get,

A =

(vi) Given a_{i j} = ½ |-3i + j|

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ =
a₁₂ =
a₁₃ =
a₁₄ =
a₂₁ =

a₂₂ =

a₂₃ =

a_{24 =}
a₃₁ =
a₃₂ =
a₃₃ =
a₃₄ =

Substituting these values in matrix A we get,

A =

Multiplying by negative sign we get,

7. Construct a 4 × 3 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) a_{i j} = 2i + i/j

(ii) a_{i j} = (i – j)/ (i + j)

(iii) a_{i j} = i

Solution:

(i) Given a_{i j} = 2i + i/j

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A we get,

A =

(ii) Given a_{i j} = (i – j)/ (i + j)

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A we get,

A =

(iii) Given a_{i j} = i

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ = 1

a₁₂ = 1

a₁₃ = 1

a₂₁ = 2

a₂₂ = 2

a₂₃ = 2

a₃₁ = 3

a₃₂ = 3

a₃₃ = 3

a₄₁ = 4

a₄₂ = 4

a₄₃ = 4

Substituting these values in matrix A we get,

A =

8. Find x, y, a and b if

Solution:

Given

Given that two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

Therefore by equating them we get,

3x + 4y = 2 …… (1)

x – 2y = 4 …… (2)

a + b = 5 …… (3)

2a – b = – 5 …… (4)

Multiplying equation (2) by 2 and adding to equation (1), we get

3x + 4y + 2x – 4y = 2 + 8

⇒ 5x = 10

⇒ x = 2

Now, substituting the value of x in equation (1)

3 × 2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Now by adding equation (3) and (4)

a + b + 2a – b = 5 + (– 5)

⇒ 3a = 5 – 5 = 0

⇒ a = 0

Now, again by substituting the value of a in equation (3), we get

0 + b = 5

⇒ b = 5

∴ a = 0, b = 5, x = 2 and y = – 1

9. Find x, y, a and b if

Solution:

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

10. Find the values of a, b, c and d from the following equations:

Solution:

Given

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

---

Chapter 5 Algebra of Matrices Ex 5.2

1. Compute the following sums:

Solution:

(i) Given

Corresponding elements of two matrices should be added

Therefore, we get

Therefore,

(ii) Given

Therefore,

Find each of the following:

(i) 2A – 3B

(ii) B – 4C

(iii) 3A – C

(iv) 3A – 2B + 3C

Solution:

(i) Given

First we have to compute 2A

Now by computing 3B we get,

Now by we have to compute 2A – 3B we get

Therefore

(ii) Given

First we have to compute 4C,

Now,

Therefore we get,

(iii) Given

First we have to compute 3A,

Now,

Therefore,

(iv) Given

First we have to compute 3A

Now we have to compute 2B

By computing 3C we get,

Therefore,

(i) A + B and B + C

(ii) 2B + 3A and 3C – 4B

Solution:

(i) Consider A + B,

A + B is not possible because matrix A is an order of 2 x 2 and Matrix B is an order of 2 x 3, so the Sum of the matrix is only possible when their order is same.

Now consider B + C

(ii) Consider 2B + 3A

2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix.

Now consider 3C – 4B,

Solution:

Given

Now we have to compute 2A – 3B + 4C

5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find

(i) A – 2B

(ii) B + C – 2A

(iii) 2A + 3B – 5C

Solution:

(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

We have to find B + C – 2A

Here,

Now we have to compute B + C – 2A

(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

Now we have to find 2A + 3B – 5C

Here,

Now consider 2A + 3B – 5C

6. Given the matrices

Verify that (A + B) + C = A + (B + C)

Solution:

Given

Now we have to verify (A + B) + C = A + (B + C)

First consider LHS, (A + B) + C,

Now consider RHS, that is A + (B + C)

Therefore LHS = RHS

Hence (A + B) + C = A + (B + C)

7. Find the matrices X and Y,

Solution:

Consider,

Now by simplifying we get,

Therefore,

Again consider,

Now by simplifying we get,

Therefore,

Solution:

Given

Now by transposing, we get

Therefore,

Solution:

Given

Now by multiplying equation (1) and (2) we get,

Now by adding equation (2) and (3) we get,

Now by substituting X in equation (2) we get,

Solution:

Consider

Now, again consider

Therefore,

And

---

Chapter 5 Algebra of Matrices Ex 5.3

1. Compute the indicated products:

Solution:

(i) Consider

On simplification we get,

(ii) Consider

On simplification we get,

(iii) Consider

On simplification we get,

2. Show that AB ≠ BA in each of the following cases:

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(iii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

3. Compute the products AB and BA whichever exists in each of the following cases:

Solution:

(i) Consider,

BA does not exist

Because the number of columns in B is greater than the rows in A

(ii) Consider,

Again consider,

(iii) Consider,

AB = 0+(−1)+6+6

AB = 11

Again consider,

(iv) Consider,

4. Show that AB ≠ BA in each of the following cases:

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Again consider,

From equation (1) and (2) it is clear that,

AB ≠ BA

5. Evaluate the following:

Solution:

(i) Given

First we have to add first two matrix,

On simplifying, we get

(ii) Given,

First we have to multiply first two given matrix,

= 82

(iii) Given

First we have subtract the matrix which is inside the bracket,

Solution:

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equation (1), (2), (3) and (4), it is clear that A² = B²= C²= I₂

Solution:

Given

Consider,

Now we have to find,

Solution:

Given

Consider,

Hence the proof.

Solution:

Given,

Consider,

Again consider,

Hence the proof.

Solution:

Given,

Consider,

Hence the proof.

Solution:

Given,

Consider,

We know that,

Again we have,

Solution:

Given,

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_3×3

Solution:

Given

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_3×3

Solution:

Given

Now consider,

Therefore AB = A

Again consider, BA we get,

Hence BA = B

Hence the proof.

Solution:

Given,

Consider,

Now again consider, B²

Now by subtracting equation (2) from equation (1) we get,

16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)

Solution:

(i) Given

Consider,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

(ii) Given,

Consider the LHS,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

Solution:

(i) Given

Consider LHS,

Now consider RHS,

From equation (1) and (2), it is clear that A (B + C) = AB + AC

(ii) Given,

Consider the LHS

Now consider RHS,

Solution:

Given,

Consider the LHS,

Now consider RHS

From the above equations LHS = RHS

Therefore, A (B – C) = AB – AC.

19. Compute the elements a₄₃ and a₂₂ of the matrix:

Solution:

Given

From the above matrix, a₄₃ = 8and a₂₂ = 0

Solution:

Given

Consider,

Again consider,

Now, consider the RHS

Therefore, A³ = p I + q A + rA²

Hence the proof.

21. If ω is a complex cube root of unity, show that

Solution:

Given

It is also given that ω is a complex cube root of unity,

Consider the LHS,

We know that 1 + ω + ω² = 0 and ω³ = 1

Now by simplifying we get,

Again by substituting 1 + ω + ω² = 0 and ω³ = 1 in above matrix we get,

Therefore LHS = RHS

Hence the proof.

Solution:

Given,

Consider A²

Therefore A² = A

Solution:

Given

Consider A²,

Hence A² = I₃

Solution:

(i) Given

= 2x+1+2+x+3 = 0

= 3x+6 = 0

= 3x = -6

x = -6/3

x = -2

(ii) Given,

On comparing the above matrix we get,

x = 13

Solution:

Given

⇒ (2x+4)x+4(x+2)–1(2x+4) = 0

⇒ 2x² + 4x + 4x + 8 – 2x – 4 = 0

⇒ 2x² + 6x + 4 = 0

⇒ 2x² + 2x + 4x + 4 = 0

⇒ 2x (x + 1) + 4 (x + 1) = 0

⇒ (x + 1) (2x + 4) = 0

⇒ x = -1 or x = -2

Hence, x = -1 or x = -2

Solution:

Given

By multiplying we get,

Solution:

Given

Now we have to prove A² – A + 2 I = 0

Solution:

Given

Solution:

Given

Hence the proof.

Solution:

Given

Hence the proof.

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

I is identity matrix, so

Also given,

Now, we have to find A², we get

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 7

Solution:

Given

To show that f (A) = 0

Substitute x = A in f(x), we get

I is identity matrix, so

Now, we will find the matrix for A², we get

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get

So,

Hence Proved

Solution:

Given

So

Now, we will find the matrix for A², we get

Now, we will find the matrix for λ A, we get

But given, A² = λ A + μ I

Substitute corresponding values from equation (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

39. Find the value of x for which the matrix product

Solution:

We know,

is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_in b_nj, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is

---

Chapter 5 Algebra of Matrices Ex 5.4

(i) (2A)^T = 2 A^T

(ii) (A + B)^T = A^T + B^T

(iii) (A – B)^T = A^T – B^T

(iv) (AB)^T = B^T A^T

Solution:

(i) Given

Consider,

Put the value of A

L.H.S = R.H.S

(ii) Given

Consider,

L.H.S = R.H.S

Hence proved.

(iii) Given

Consider,

L.H.S = R.H.S

(iv) Given

So,

Solution:

Given

L.H.S = R.H.S

So,

(i) A + B)^T = A^T + B^T

(ii) (AB)^T = B^T A^T

(iii) (2A)^T = 2 A^T

Solution:

(i) Given

Consider,

L.H.S = R.H.S

So,

(ii) Given

Consider,

L.H.S = R.H.S

So,

(iii) Given

Consider,

L.H.S = R.H.S

So,

Solution:

Given

Consider,

L.H.S = R.H.S

So,

Solution:

Given

Now we have to find (AB)^T

So,

---

Chapter 5 Algebra of Matrices Ex 5.5

Solution:

Given

Consider,

 … (i)

 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric.

Solution:

Given

Consider,

 … (i)

 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric matrix.

Solution:

Given,

 is a symmetric matrix.

We know that A = [a_ij]_{m × n} is a symmetric matrix if a_ij = a_ji

So,

Hence, x = 4, y = 2, t = -3 and z can have any value.

4. Let. Find matrices X and Y such that X + Y = A, where X is a symmetric and y is a skew-symmetric matrix.

Solution:

Given,
 Then

Now,

Now,

X is a symmetric matrix.

Now,

-Y ^T = Y

Y is a skew symmetric matrix.

Hence, X + Y = A

Chapter 3 Binary Operations Ex 3.1

1. Determine whether the following operation define a binary operation on the given set or not:

(i) ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

(ii) ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

(iii)  ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

(iv) ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×₆ b = Remainder when a b is divided by 6.

(v) ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a +₆ b

(vi) ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

Solution:

(i) Given ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

Let a, b ∈ N. Then,

a^b ∈ N      [∵ a^b≠0 and a, b is positive integer]

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

(ii) Given ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

Both a = 3 and b = -1 belong to Z.

⇒ a * b = 3^-1

= 1/3 ∉ Z

Thus, * is not a binary operation on Z.

(iii)  Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

So, * is not a binary operation on N.

(iv) Given ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×₆ b = Remainder when a b is divided by 6.

Consider the composition table,

X6	1	2	3	4	5
1	1	2	3	4	5
2	2	4	0	2	4
3	3	0	3	0	3
4	4	2	0	4	2
5	5	4	3	2	1

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ S

Thus, ×₆ is not a binary operation on S.

(v) Given ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a +₆ b

Consider the composition table,

+6	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(vi) Given ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

a^b, b^a ∈ N

⇒ a^b + b^a ∈ N      ∵AdditionisbinaryoperationonN

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii) Given ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = (a – 1)/ (b + 1)

= (2 – 1)/ (- 1 + 1)

= 1/0 whichisnotdefined

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z⁺, defined * by a * b = a – b

(ii) On Z⁺, define * by a*b = ab

(iii) On R, define * by a*b = ab²

(iv) On Z⁺ define * by a * b = |a − b|

(v) On Z⁺ define * by a * b = a

(vi) On R, define * by a * b = a + 4b²

Here, Z⁺ denotes the set of all non-negative integers.

Solution:

(i) Given On Z⁺, defined * by a * b = a – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

= -1 ∉ Z⁺ [because Z⁺ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z⁺

Thus, * is not a binary operation on Z⁺.

(ii) Given Z⁺, define * by a*b = a b

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ | a – b | ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(v) Given on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(vi) Given On R, define * by a * b = a + 4b²

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ R

Thus, * is a binary operation on R.

3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.

Solution:

Given a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

4. Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.

Solution:

LCM	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	5	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 ∉ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

5. Let S = {a, b, c}. Find the total number of binary operations on S.

Solution:

Number of binary operations on a set with n elements is (n^{n^{2}})

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is (3^{3^{2}})

Chapter 3 Binary Operations Ex 3.2

1. Let ‘*’ be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b ∈ N
(i) Find 2 * 4, 3 * 5, 1 * 6.

(ii) Check the commutativity and associativity of ‘*’ on N.

Solution:

(i) Given a * b = 1.c.m. (a, b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5)

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) We have to prove commutativity of *

Let a, b ∈ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

Therefore

a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c ) = a * l.c.m. (b, c)

= l.c.m. (a, (b, c))

= l.c.m (a, b, c)

(a * b) * c = l.c.m. (a, b) * c

= l.c.m. ((a, b), c)

= l.c.m. (a, b, c)

Therefore

(a * (b * c) = (a * b) * c, ∀ a, b , c ∈ N

Thus, * is associative on N.

2. Determine which of the following binary operation is associative and which is commutative:

(i) * on N defined by a * b = 1 for all a, b ∈ N

(ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q

Solution:

(i) We have to prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

Then a * (b * c) = a * (1)

= 1

(a * b) *c = (1) * c

= 1

Therefore a * (b * c) = (a * b) *c for all a, b, c ∈ N

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

Let a, b ∈ N

a * b = (a + b)/2

= (b + a)/2

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c) = a * (b + c)/2

= a+(b+c)/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= (a+b)/2+c /2

= (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b= 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2

= 1 * (5/2)

= 1+(5/2)/2

= 7/4

(1 * 2) * 3 = (1 + 2)/2 * 3

= 3/2 * 3

= (3/2)+3/2

= 4/9

Therefore, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

3. Let A be any set containing more than one element. Let ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?

Solution:

Let a, b ∈ A

Then, a * b = b

b * a = a

Therefore a * b ≠ b * a

Thus, * is not commutative on A

Now we have to check associativity:

Let a, b, c ∈ A

a * (b * c) = a * c

= c

Therefore

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Thus, * is associative on A

4. Check the commutativity and associativity of each of the following binary operations:

(i) ‘*’ on Z defined by a * b = a + b + a b for all a, b ∈ Z 

(ii) ‘*’ on N defined by a * b = 2^ab for all a, b ∈ N

(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q

(iv) ‘⊙’ on Q defined by a ⊙ b = a² + b² for all a, b ∈ Q

(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q

(vi) ‘*’ on Q defined by a * b = ab² for all a, b ∈ Q

(vii) ‘*’ on Q defined by a * b = a + a b for all a, b ∈ Q

(viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R

(ix) ‘*’ on Q defined by a * b = (a – b)² for all a, b ∈ Q

(x) ‘*’ on Q defined by a * b = a b + 1 for all a, b ∈ Q

(xi) ‘*’ on N defined by a * b = a^b for all a, b ∈ N

(xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z

(xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q

(xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z

(xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q

Solution:

(i) First we have to check commutativity of *

Let a, b ∈ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Now we have to prove associativity of *

Let a, b, c ∈ Z, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) First we have to check commutativity of *

Let a, b ∈ N

a * b = 2^ab

= 2^ba

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

Now we have to check associativity of *

Let a, b, c ∈ N

Then, a * (b * c) = a * (2^bc)

=(2^{a*2^{bc}})

(a * b) * c = (2^ab) * c

=(2^{ab*2^{c}})

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(iii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Therefore, a * b ≠ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Therefore,

a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q

(iv) First we have to check commutativity of ⊙

Let a, b ∈ Q, then

a ⊙ b = a² + b²

= b² + a²

= b ⊙ a

Therefore, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ on Q

Now we have to check associativity of ⊙

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)²

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

Therefore,

(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) First we have to check commutativity of o

Let a, b ∈ Q, then

a o b = (ab/2)

= (b a/2)

= b o a

Therefore, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q

Now we have to check associativity of o

Let a, b, c ∈ Q, then

a o (b o c) = a o (b c/2)

= a(bc/2)/2

= a(bc/2)/2

= (a b c)/4

(a o b) o c = (ab/2) o c

= (ab/2)c /2

= (a b c)/4

Therefore a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab²

b * a = ba²

Therefore,

a * b ≠ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc²)

= a (bc²)²

= ab² c⁴

(a * b) * c = (ab²) * c

= ab²c²

Therefore a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(vii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

Therefore, a * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to prove associativity on Q.

Let a, b, c ∈ Q, then

a * (b * c) = a * (b + b c)

= a + a (b + b c)

= a + ab + a b c

(a * b) * c = (a + a b) * c

= (a + a b) + (a + a b) c

= a + a b + a c + a b c

Therefore a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii) First we have to check commutativity of *

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7

= b * a

Therefore,

a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R

Now we have to prove associativity of * on R.

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

(a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

Therefore,

a * (b * c ) = (a * b) * c, for all a, b, c ∈ R

Thus, * is associative on R.

(ix) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = (a – b)²

= (b – a)²

= b * a

Therefore,

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)²

= a * (b² + c² – 2 b c)

= (a – b² – c² + 2bc)²

(a * b) * c = (a – b)² * c

= (a² + b² – 2ab) * c

= (a² + b² – 2ab – c)²

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc + 1)

= a (b c + 1) + 1

= a b c + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi) First we have to check commutativity of *

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

Therefore, a * b ≠ b * a

Thus, * is not commutative on N.

Now we have to check associativity of *

a * (b * c) = a * (b^c)

=

(a * b) * c = (a^b) * c

= (a^b)^c

= a^bc

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(xii) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Therefore,

a * b ≠ b * a

Thus, * is not commutative on Z.

Now we have to check associativity of *

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z

(xiii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b c/4)

= a(bc/4)/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= (ab/4)c/4

= a b c/16

Therefore,

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Thus, * is associative on Q.

(xiv) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

Therefore, a * b = b * a, for all a, b ∈ Z

Thus, * is commutative on Z.

Now we have to check associativity of *

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – b c)

= a + b + c- b c – ab – ac + a b c

(a * b) * c = (a + b – a b) c

= a + b – ab + c – (a + b – ab) c

= a + b + c – ab – ac – bc + a b c

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(xv) First we have to check commutativity of *

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

Therefore, a * b = b * a, for all a, b ∈ N

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ N

a * (b * c) = a * gcd(a,b)

= gcd (a, b, c)

(a * b) * c = gcd(a,b) * c

= gcd (a, b, c)

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ N

Thus, * is associative on N.

5. If the binary operation o is defined by a0b = a + b – ab on the set Q – {-1} of all rational numbers other than 1, show that o is commutative on Q – 1.

Solution:

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – ba

= boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}

6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?

Solution:

Let a, b ∈ Z

a * b = 3a + 7b

b * a = 3b + 7a

Thus, a * b ≠ b * a

Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2

= 3 + 14

= 17

2 * 1 = 3 × 2 + 7 × 1

= 6 + 7

= 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

7. On the set Z of integers a binary operation * is defined by a 8 b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.

Solution:

Let a, b, c ∈ Z

a * (b * c) = a * (bc + 1)

= a (bc + 1) + 1

= a b c + a + 1

(a * b) * c = (ab+ 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Thus, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

Chapter 3 Binary Operations Ex 3.3

1. Find the identity element in the set I⁺ of all positive integers defined by a * b = a + b for all a, b ∈ I⁺.

Solution:

Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ I⁺

a * e = a and e * a = a, ∀ a ∈ I⁺

a + e = a and e + a = a, ∀ a ∈ I⁺

e = 0, ∀ a ∈ I⁺

Thus, 0 is the identity element in I⁺ with respect to *.

2. Find the identity element in the set of all rational numbers except – 1 with respect to * defined by a * b = a + b + ab

Solution:

Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} becauseanotequalto−1

Thus, 0 is the identity element in Q – {-1} with respect to *.

---

Chapter 3 Binary Operations Ex 3.4

1. Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z.

(i) Show that * is both commutative and associative.

(ii) Find the identity element in Z

(iii) Find the invertible element in Z.

Solution:

(i) First we have to prove commutativity of *

Let a, b ∈ Z. then,

a * b = a + b – 4

= b + a – 4

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Thus, * is commutative on Z.

Now we have to prove associativity of Z.

Let a, b, c ∈ Z. then,

a * (b * c) = a * (b + c – 4)

= a + b + c -4 – 4

= a + b + c – 8

(a * b) * c = (a + b – 4) * c

= a + b – 4 + c – 4

= a + b + c – 8

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that

a * e = a = e * a ∀ a ∈ Z

a * e = a and e * a = a, ∀ a ∈ Z

a + e – 4 = a and e + a – 4 = a, ∀ a ∈ Z

e = 4, ∀ a ∈ Z

Thus, 4 is the identity element in Z with respect to *.

(iii) Let a ∈ Z and b ∈ Z be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b – 4 = 4 and b + a – 4 = 4

b = 8 – a ∈ Z

Thus, 8 – a is the inverse of a ∈ Z

2. Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by a * b = (3ab/5) for all a, b ∈ Q₀. Show that * is commutative as well as associative. Also, find its identity element, if it exists.

Solution:

First we have to prove commutativity of *

Let a, b ∈ Q₀

a * b = (3ab/5)

= (3ba/5)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q₀

Now we have to prove associativity of *

Let a, b, c ∈ Q₀

a * (b * c) = a * (3bc/5)

= a(3bc/5) /5

= 3 abc/25

(a * b) * c = (3 ab/5) * c

= (3ab/5)c/ 5

= 3 abc /25

Therefore a * (b * c) = (a * b) * c, for all a, b, c ∈ Q₀

Thus * is associative on Q₀

Now we have to find the identity element

Let e be the identity element in Z with respect to * such that

a * e = a = e * a ∀ a ∈ Q₀

a * e = a and e * a = a, ∀ a ∈ Q₀

3ae/5 = a and 3ea/5 = a, ∀ a ∈ Q₀

e = 5/3 ∀ a ∈ Q₀ becauseaisnotequalto0

Thus, 5/3 is the identity element in Q₀ with respect to *.

3. Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then,

(i) Show that * is both commutative and associative on Q – {-1}

(ii) Find the identity element in Q – {-1}

(iii) Show that every element of Q – {-1} is invertible. Also, find inverse of an arbitrary element.

Solution:

(i) First we have to check commutativity of *

Let a, b ∈ Q – {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Q – {-1}

Now we have to prove associativity of *

Let a, b, c ∈ Q – {-1}, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1}

Thus, * is associative on Q – {-1}.

(ii) Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} becauseanotequalto−1

Thus, 0 is the identity element in Q – {-1} with respect to *.

(iii) Let a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b + ab = 0 and b + a + ba = 0

b (1 + a) = – a Q – {-1}

b = -a/1 + a Q – {-1} becauseanotequalto−1

Thus, -a/1 + a is the inverse of a ∈ Q – {-1}

4. Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.

(i) Show that ‘O’ is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A.

Solution:

(i) Let X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

Therefore,

X O Y = Y O X, ∀ X, Y ∈ A

Thus, O is not commutative on A.

Now we have to check associativity of O

Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R₀ and b, d, f ∈ R

X O (Y O Z) = (a, b) O (ce, de + f)

= (ace, bce + de + f)

(X O Y) O Z = (ac, bc + d) O (e, f)

= (ace, (bc + d) e + f)

= (ace, bce + de + f)

Therefore, X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A

(ii) Let E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R

Such that,

X O E = X = E O X, ∀ X ∈ A

X O E = X and EOX = X

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)

Considering (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 sincex=1

Considering (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 sincex=1

Therefore (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Considering (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a sincem=1/a

Considering (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

Therefore the inverse of (a, b) ∈ A with respect to O is (1/a, -b/a)

Chapter 3 Binary Operations Ex 3.5

1. Construct the composition table for ×₄ on set S = {0, 1, 2, 3}.

Solution:

Given that ×₄ on set S = {0, 1, 2, 3}

Here,

1 ×₄ 1 = remainder obtained by dividing 1 × 1 by 4

= 1

0 ×₄ 1 = remainder obtained by dividing 0 × 1 by 4

= 0

2 ×₄ 3 = remainder obtained by dividing 2 × 3 by 4

= 2

3 ×₄ 3 = remainder obtained by dividing 3 × 3 by 4

= 1

So, the composition table is as follows:

×4	0	1	2	3
0	0	0	0	0
1	0	1	2	3
2	0	2	0	2
3	0	3	2	1

2. Construct the composition table for +₅ on set S = {0, 1, 2, 3, 4}

Solution:

1 +₅ 1 = remainder obtained by dividing 1 + 1 by 5

= 2

3 +₅ 1 = remainder obtained by dividing 3 + 1 by 5

= 2

4 +₅ 1 = remainder obtained by dividing 4 + 1 by 5

= 3

So, the composition table is as follows:

+5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

3. Construct the composition table for ×₆ on set S = {0, 1, 2, 3, 4, 5}.

Solution:

Here,

1 ×₆ 1 = remainder obtained by dividing 1 × 1 by 6

= 1

3 ×₆ 4 = remainder obtained by dividing 3 × 4 by 6

= 0

4 ×₆ 5 = remainder obtained by dividing 4 × 5 by 6

= 2

So, the composition table is as follows:

×6	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	1	2	3	4	5
2	0	2	4	0	2	4
3	0	3	0	3	0	3
4	0	4	2	0	4	2
5	0	5	4	3	2	1

4. Construct the composition table for ×₅ on set Z₅ = {0, 1, 2, 3, 4}

Solution:

Here,

1 ×₅ 1 = remainder obtained by dividing 1 × 1 by 5

= 1

3 ×₅ 4 = remainder obtained by dividing 3 × 4 by 5

= 2

4 ×₅ 4 = remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×5	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

5. For the binary operation ×₁₀ set S = {1, 3, 7, 9}, find the inverse of 3.

Solution:

Here,

1 ×₁₀ 1 = remainder obtained by dividing 1 × 1 by 10

= 1

3 ×₁₀ 7 = remainder obtained by dividing 3 × 7 by 10

= 1

7 ×₁₀ 9 = remainder obtained by dividing 7 × 9 by 10

= 3

So, the composition table is as follows:

×10	1	3	7	9
1	1	3	7	9
3	3	9	1	7
7	7	1	9	3
9	9	7	3	1

From the table we can observe that elements of first row as same as the top-most row.

So, 1 ∈ S is the identity element with respect to ×₁₀

Now we have to find inverse of 3

3 ×₁₀ 7 = 1

So the inverse of 3 is 7.

Chapter 4 Inverse Trigonometric Functions Ex 4.1

Q1. Find the principal value of the following:

Solution:

(iii) Given functions can be written as

(iv) The given question can be written as

(v) Let

(vi) Let

2.

(i) 

(ii) 

Solution:

(i) The given question can be written as,

(ii) Given question can be written as

Chapter 4 Inverse Trigonometric Functions Ex 4.2

Q1. Find the domain of definition of f(x) = cos ^-1 (x² – 4)

Solution:

Given f(x) = cos ^-1 (x² – 4)

We know that domain of cos^-1 (x² – 4) lies in the interval −1,1

Therefore, we can write as

-1 ≤ x² – 4 ≤ 1

4 – 1 ≤ x² ≤ 1 + 4

3 ≤ x² ≤ 5

±√ 3 ≤ x ≤ ±√5

– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5

Therefore domain of cos^-1 (x² – 4) is −√5,–√3 ∪ √3,√5

Q2. Find the domain of f(x) = cos^-1 2x + sin^-1 x.

Solution:

Given that f(x) = cos^-1 2x + sin^-1 x.

Now we have to find the domain of f(x),

We know that domain of cos^-1 x lies in the interval −1,1

Also know that domain of sin^-1 x lies in the interval −1,1

Therefore, the domain of cos^-1 (2x) lies in the interval −1,1

Hence we can write as,

-1 ≤ 2x ≤ 1

– ½ ≤ x ≤ ½

Hence, domain of cos^-1(2x) + sin^-1 x lies in the interval ½½−½,½

Chapter 4 Inverse Trigonometric Functions Ex 4.3

Q1. Find the principal value of each of the following:

(i) tan^-1 (1/√3)

(ii) tan^-1 (-1/√3)

(iii) tan^-1 (cos (π/2))

(iv) tan^-1 (2 cos (2π/3))

Solution:

(i) Given tan^-1 (1/√3)

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to π/6

Therefore tan^-1 (1/√3) = π/6

Hence the principal value of tan^-1 (1/√3) = π/6

(ii) Given tan^-1 (-1/√3)

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (-1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to -π/6

Therefore tan^-1 (-1/√3) = -π/6

Hence the principal value of tan^-1 (-1/√3) = – π/6

(iii) Given that tan^-1 (cos (π/2))

But we know that cos (π/2) = 0

We know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore tan^-1 (0) = 0

Hence the principal value of tan^-1 (cos (π/2) is 0.

(iv) Given that tan^-1 (2 cos (2π/3))

But we know that cos π/3 = 1/2

So, cos (2π/3) = -1/2

Therefore tan^-1 (2 cos (2π/3)) = tan^-1 (2 × – ½)

= tan^-1(-1)

= – π/4

Hence, the principal value of tan^-1 (2 cos (2π/3)) is – π/4

Chapter 4 Inverse Trigonometric Functions Ex 4.4

Q1. Find the principal value of each of the following:

(i) sec^-1 (-√2)

(ii) sec^-1 (2)

(iii) sec^-1 (2 sin (3π/4))

(iv) sec^-1 (2 tan (3π/4))

Solution:

(i) Given sec^-1 (-√2)

Now let y = sec^-1 (-√2)

Sec y = -√2

We know that sec π/4 = √2

Therefore, -sec (π/4) = -√2

= sec (π – π/4)

= sec (3π/4)

Thus the range of principal value of sec^-1 is 0,π – {π/2}

And sec (3π/4) = – √2

Hence the principal value of sec^-1 (-√2) is 3π/4

(ii) Given sec^-1 (2)

Let y = sec^-1 (2)

Sec y = 2

= Sec π/3

Therefore the range of principal value of sec^-1 is 0,π – {π/2} and sec π/3 = 2

Thus the principal value of sec^-1 (2) is π/3

(iii) Given sec^-1 (2 sin (3π/4))

But we know that sin (3π/4) = 1/√2

Therefore 2 sin (3π/4) = 2 × 1/√2

2 sin (3π/4) = √2

Therefore by substituting above values in sec^-1 (2 sin (3π/4)), we get

Sec^-1 (√2)

Let Sec^-1 (√2) = y

Sec y = √2

Sec (π/4) = √2

Therefore range of principal value of sec^-1 is 0,π – {π/2} and sec (π/4) = √2

Thus the principal value of sec^-1 (2 sin (3π/4)) is π/4.

(iv) Given sec^-1 (2 tan (3π/4))

But we know that tan (3π/4) = -1

Therefore, 2 tan (3π/4) = 2 × -1

2 tan (3π/4) = -2

By substituting these values in sec^-1 (2 tan (3π/4)), we get

Sec^-1 (-2)

Now let y = Sec^-1 (-2)

Sec y = – 2

– sec (π/3) = -2

= sec (π – π/3)

= sec (2π/3)

Therefore the range of principal value of sec^-1 is 0,π – {π/2} and sec (2π/3) = -2

Thus, the principal value of sec^-1 (2 tan (3π/4)) is (2π/3).

Chapter 4 Inverse Trigonometric Functions Ex 4.5

Q1. Find the principal values of each of the following:

(i) cosec^-1 (-√2)

(ii) cosec^-1 (-2)

(iii) cosec^-1 (2/√3)

(iv) cosec^-1 (2 cos (2π/3))

Solution:

(i) Given cosec^-1 (-√2)

Let y = cosec^-1 (-√2)

Cosec y = -√2

– Cosec y = √2

– Cosec (π/4) = √2

– Cosec (π/4) = cosec (-π/4) since–cosecθ=cosec(−θ)

The range of principal value of cosec^-1 −π/2,π/2 – {0} and cosec (-π/4) = – √2

Cosec (-π/4) = – √2

Therefore the principal value of cosec^-1 (-√2) is – π/4

(ii) Given cosec^-1 (-2)

Let y = cosec^-1 (-2)

Cosec y = -2

– Cosec y = 2

– Cosec (π/6) = 2

– Cosec (π/6) = cosec (-π/6) since–cosecθ=cosec(−θ)

The range of principal value of cosec^-1 −π/2,π/2 – {0} and cosec (-π/6) = – 2

Cosec (-π/6) = – 2

Therefore the principal value of cosec^-1 (-2) is – π/6

(iii) Given cosec^-1 (2/√3)

Let y = cosec^-1 (2/√3)

Cosec y = (2/√3)

Cosec (π/3) = (2/√3)

Therefore range of principal value of cosec^-1 is −π/2,π/2 – {0} and cosec (π/3) = (2/√3)

Thus, the principal value of cosec^-1 (2/√3) is π/3

(iv) Given cosec^-1 (2 cos (2π/3))

But we know that cos (2π/3) = – ½

Therefore 2 cos (2π/3) = 2 × – ½

2 cos (2π/3) = -1

By substituting these values in cosec^-1 (2 cos (2π/3)) we get,

Cosec^-1 (-1)

Let y = cosec^-1 (-1)

– Cosec y = 1

– Cosec (π/2) = cosec (-π/2) since–cosecθ=cosec(−θ)

The range of principal value of cosec^-1 −π/2,π/2 – {0} and cosec (-π/2) = – 1

Cosec (-π/2) = – 1

Therefore the principal value of cosec^-1 (2 cos (2π/3)) is – π/2

Chapter 4 Inverse Trigonometric Functions Ex 4.6

Q1. Find the principal values of each of the following:

(i) cot^-1(-√3)

(ii) Cot^-1(√3)

(iii) cot^-1(-1/√3)

(iv) cot^-1(tan 3π/4)

Solution:

(i) Given cot^-1(-√3)

Let y = cot^-1(-√3)

– Cot (π/6) = √3

= Cot (π – π/6)

= cot (5π/6)

The range of principal value of cot^-1 is (0, π) and cot (5 π/6) = – √3

Thus, the principal value of cot^-1 (- √3) is 5π/6

(ii) Given Cot^-1(√3)

Let y = cot^-1(√3)

Cot (π/6) = √3

The range of principal value of cot^-1 is (0, π) and

Thus, the principal value of cot^-1 (√3) is π/6

(iii) Given cot^-1(-1/√3)

Let y = cot^-1(-1/√3)

Cot y = (-1/√3)

– Cot (π/3) = 1/√3

= Cot (π – π/3)

= cot (2π/3)

The range of principal value of cot^-1(0, π) and cot (2π/3) = – 1/√3

Therefore the principal value of cot^-1(-1/√3) is 2π/3

(iv) Given cot^-1(tan 3π/4)

But we know that tan 3π/4 = -1

By substituting this value in cot^-1(tan 3π/4) we get

Cot^-1(-1)

Now, let y = cot^-1(-1)

Cot y = (-1)

– Cot (π/4) = 1

= Cot (π – π/4)

= cot (3π/4)

The range of principal value of cot^-1(0, π) and cot (3π/4) = – 1

Therefore the principal value of cot^-1(tan 3π/4) is 3π/4

Chapter 4 Inverse Trigonometric Functions Ex 4.7

Q1. Evaluate each of the following:

(i) sin^-1(sin π/6)

(ii) sin^-1(sin 7π/6)

(iii) sin^-1(sin 5π/6)

(iv) sin^-1(sin 13π/7)

(v) sin^-1(sin 17π/8)

(vi) sin^-1{(sin – 17π/8)}

(vii) sin^-1(sin 3)

(viii) sin^-1(sin 4)

(ix) sin^-1(sin 12)

(x) sin^-1(sin 2)

Solution:

(i) Given sin^-1(sin π/6)

We know that the value of sin π/6 is ½

By substituting this value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Now let y = sin^-1 (1/2)

Sin (π/6) = ½

The range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

Therefore sin^-1(sin π/6) = π/6

(ii) Given sin^-1(sin 7π/6)

But we know that sin 7π/6 = – ½

By substituting this in sin^-1(sin 7π/6) we get,

Sin^-1 (-1/2)

Now let y = sin^-1 (-1/2)

– Sin y = ½

– Sin (π/6) = ½

– Sin (π/6) = sin (- π/6)

The range of principal value of sin^-1(-π/2, π/2) and sin (- π/6) = – ½

Therefore sin^-1(sin 7π/6) = – π/6

(iii) Given sin^-1(sin 5π/6)

We know that the value of sin 5π/6 is ½

By substituting this value in sin^-1(sin 5π/6)

We get, sin^-1 (1/2)

Now let y = sin^-1 (1/2)

Sin (π/6) = ½

The range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

Therefore sin^-1(sin 5π/6) = π/6

(iv) Given sin^-1(sin 13π/7)

Given question can be written as sin (2π – π/7)

Sin (2π – π/7) can be written as sin (-π/7) sincesin(2π–θ)=sin(−θ)

By substituting these values in sin^-1(sin 13π/7) we get sin^-1(sin – π/7)

As sin^-1(sin x) = x with x ∈ −π/2,π/2

Therefore sin^-1(sin 13π/7) = – π/7

(v) Given sin^-1(sin 17π/8)

Given question can be written as sin (2π + π/8)

Sin (2π + π/8) can be written as sin (π/8)

By substituting these values in sin^-1(sin 17π/8) we get sin^-1(sin π/8)

As sin^-1(sin x) = x with x ∈ −π/2,π/2

Therefore sin^-1(sin 17π/8) = π/8

(vi) Given sin^-1{(sin – 17π/8)}

But we know that – sin θ = sin (-θ)

Therefore (sin -17π/8) = – sin 17π/8

– Sin 17π/8 = – sin (2π + π/8) sincesin(2π–θ)=−sin(θ)

It can also be written as – sin (π/8)

– Sin (π/8) = sin (-π/8) since–sinθ=sin(−θ)

By substituting these values in sin^-1{(sin – 17π/8)} we get,

Sin^-1(sin – π/8)

As sin^-1(sin x) = x with x ∈ −π/2,π/2

Therefore sin^-1(sin -π/8) = – π/8

(vii) Given sin^-1(sin 3)

We know that sin^-1(sin x) = x with x ∈ −π/2,π/2 which is approximately equal to −1.57,1.57

But here x = 3, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 3) = sin (3) also π – 3 ∈ −π/2,π/2

Sin^-1(sin 3) = π – 3

(viii) Given sin^-1(sin 4)

We know that sin^-1(sin x) = x with x ∈ −π/2,π/2 which is approximately equal to −1.57,1.57

But here x = 4, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 4) = sin (4) also π – 4 ∈ −π/2,π/2

Sin^-1(sin 4) = π – 4

(ix) Given sin^-1(sin 12)

We know that sin^-1(sin x) = x with x ∈ −π/2,π/2 which is approximately equal to −1.57,1.57

But here x = 12, which does not lie on the above range,

Therefore we know that sin (2nπ – x) = sin (-x)

Hence sin (2nπ – 12) = sin (-12)

Here n = 2 also 12 – 4π ∈ −π/2,π/2

Sin^-1(sin 12) = 12 – 4π

(x) Given sin^-1(sin 2)

We know that sin^-1(sin x) = x with x ∈ −π/2,π/2 which is approximately equal to −1.57,1.57

But here x = 2, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 2) = sin (2) also π – 2 ∈ −π/2,π/2

Sin^-1(sin 2) = π – 2

Q2. Evaluate each of the following:

(i) cos^-1{cos (-π/4)}

(ii) cos^-1(cos 5π/4)

(iii) cos^-1(cos 4π/3)

(iv) cos^-1(cos 13π/6)

(v) cos^-1(cos 3)

(vi) cos^-1(cos 4)

(vii) cos^-1(cos 5)

(viii) cos^-1(cos 12)

Solution:

(i) Given cos^-1{cos (-π/4)}

We know that cos (-π/4) = cos (π/4) [since cos (-θ) = cos θ

Also know that cos (π/4) = 1/√2

By substituting these values in cos^-1{cos (-π/4)} we get,

Cos^-1(1/√2)

Now let y = cos^-1(1/√2)

Therefore cos y = 1/√2

Hence range of principal value of cos^-1 is 0,π and cos (π/4) = 1/√2

Therefore cos^-1{cos (-π/4)} = π/4

(ii) Given cos^-1(cos 5π/4)

But we know that cos (5π/4) = -1/√2

By substituting these values in cos^-1{cos (5π/4)} we get,

Cos^-1(-1/√2)

Now let y = cos^-1(-1/√2)

Therefore cos y = – 1/√2

– Cos (π/4) = 1/√2

Cos (π – π/4) = – 1/√2

Cos (3 π/4) = – 1/√2

Hence range of principal value of cos^-1 is 0,π and cos (3π/4) = -1/√2

Therefore cos^-1{cos (5π/4)} = 3π/4

(iii) Given cos^-1(cos 4π/3)

But we know that cos (4π/3) = -1/2

By substituting these values in cos^-1{cos (4π/3)} we get,

Cos^-1(-1/2)

Now let y = cos^-1(-1/2)

Therefore cos y = – 1/2

– Cos (π/3) = 1/2

Cos (π – π/3) = – 1/2

Cos (2π/3) = – 1/2

Hence range of principal value of cos^-1 is 0,π and cos (2π/3) = -1/2

Therefore cos^-1{cos (4π/3)} = 2π/3

(iv) Given cos^-1(cos 13π/6)

But we know that cos (13π/6) = √3/2

By substituting these values in cos^-1{cos (13π/6)} we get,

Cos^-1(√3/2)

Now let y = cos^-1(√3/2)

Therefore cos y = √3/2

Cos (π/6) = √3/2

Hence range of principal value of cos^-1 is 0,π and cos (π/6) = √3/2

Therefore cos^-1{cos (13π/6)} = π/6

(v) Given cos^-1(cos 3)

We know that cos^-1(cos θ) = θ if 0 ≤ θ ≤ π

Therefore by applying this in given question we get,

Cos^-1(cos 3) = 3, 3 ∈ 0,π

(vi) Given cos^-1(cos 4)

We have cos^–1(cos x) = x if x ϵ 0,π ≈ 0,3.14

And here x = 4 which does not lie in the above range.

We know that cos (2π – x) = cos(x)

Thus, cos (2π – 4) = cos (4) so 2π–4 belongs in 0,π

Hence cos^–1(cos 4) = 2π – 4

(vii) Given cos^-1(cos 5)

We have cos^–1(cos x) = x if x ϵ 0,π ≈ 0,3.14

And here x = 5 which does not lie in the above range.

We know that cos (2π – x) = cos(x)

Thus, cos (2π – 5) = cos (5) so 2π–5 belongs in 0,π

Hence cos^–1(cos 5) = 2π – 5

(viii) Given cos^-1(cos 12)

Cos^–1(cos x) = x if x ϵ 0,π ≈ 0,3.14

And here x = 12 which does not lie in the above range.

We know cos (2nπ – x) = cos (x)

Cos (2nπ – 12) = cos (12)

Here n = 2.

Also 4π – 12 belongs in 0,π

∴ cos^–1(cos 12) = 4π – 12

Q3. Evaluate each of the following:

(i) tan^-1(tan π/3)

(ii) tan^-1(tan 6π/7)

(iii) tan^-1(tan 7π/6)

(iv) tan^-1(tan 9π/4)

(v) tan^-1(tan 1)

(vi) tan^-1(tan 2)

(vii) tan^-1(tan 4)

(viii) tan^-1(tan 12)

Solution:

(i) Given tan^-1(tan π/3)

As tan^-1(tan x) = x if x ϵ −π/2,π/2

By applying this condition in the given question we get,

Tan^-1(tan π/3) = π/3

(ii) Given tan^-1(tan 6π/7)

We know that tan 6π/7 can be written as (π – π/7)

Tan (π – π/7) = – tan π/7

We know that tan^-1(tan x) = x if x ϵ −π/2,π/2

Tan^-1(tan 6π/7) = – π/7

(iii) Given tan^-1(tan 7π/6)

We know that tan 7π/6 = 1/√3

By substituting this value in tan^-1(tan 7π/6) we get,

Tan^-1 (1/√3)

Now let tan^-1 (1/√3) = y

Tan y = 1/√3

Tan (π/6) = 1/√3

The range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/6) = 1/√3

Therefore tan^-1(tan 7π/6) = π/6

(iv) Given tan^-1(tan 9π/4)

We know that tan 9π/4 = 1

By substituting this value in tan^-1(tan 9π/4) we get,

Tan^-1 (1)

Now let tan^-1 (1) = y

Tan y = 1

Tan (π/4) = 1

The range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/4) = 1

Therefore tan^-1(tan 9π/4) = π/4

(v) Given tan^-1(tan 1)

But we have tan^-1(tan x) = x if x ϵ −π/2,π/2

By substituting this condition in given question

Tan^-1(tan 1) = 1

(vi) Given tan^-1(tan 2)

As tan^-1(tan x) = x if x ϵ −π/2,π/2

But here x = 2 which does not belongs to above range

We also have tan (π – θ) = –tan (θ)

Therefore tan (θ – π) = tan (θ)

Tan (2 – π) = tan (2)

Now 2 – π is in the given range

Hence tan^–1 (tan 2) = 2 – π

(vii) Given tan^-1(tan 4)

As tan^-1(tan x) = x if x ϵ −π/2,π/2

But here x = 4 which does not belongs to above range

We also have tan (π – θ) = –tan (θ)

Therefore tan (θ – π) = tan (θ)

Tan (4 – π) = tan (4)

Now 4 – π is in the given range

Hence tan^–1 (tan 2) = 4 – π

(viii) Given tan^-1(tan 12)

As tan^-1(tan x) = x if x ϵ −π/2,π/2

But here x = 12 which does not belongs to above range

We know that tan (2nπ – θ) = –tan (θ)

Tan (θ – 2nπ) = tan (θ)

Here n = 2

Tan (12 – 4π) = tan (12)

Now 12 – 4π is in the given range

∴ tan^–1 (tan 12) = 12 – 4π.

Chapter 4 Inverse Trigonometric Functions Ex 4.8

Q1. Evaluate each of the following:

(i) sin (sin^-1 7/25)

(ii) Sin (cos^-1 5/13)

(iii) Sin (tan^-1 24/7)

(iv) Sin (sec^-1 17/8)

(v) Cosec (cos^-1 8/17)

(vi) Sec (sin^-1 12/13)

(vii) Tan (cos^-1 8/17)

(viii) cot (cos^-1 3/5)

(ix) Cos (tan^-1 24/7)

Solution:

(i) Given sin (sin^-1 7/25)

Now let y = sin^-1 7/25

Sin y = 7/25 where y ∈ 0,π/2

Substituting these values in sin (sin^-1 7/25) we get

Sin (sin^-1 7/25) = 7/25

(ii) Given Sin (cos^-1 5/13)

(iii) Given Sin (tan^-1 24/7)

(iv) Given Sin (sec^-1 17/8)

(v) Given Cosec (cos^-1 8/17)

Let cos^-1(8/17) = y

cos y = 8/17 where y ∈ 0,π/2

Now, we have to find

Cosec (cos^-1 8/17) = cosec y

We know that,

sin² θ + cos² θ = 1

sin² θ = √ (1 – cos² θ)

So,

sin y = √ (1 – cos² y)

= √ (1 – (8/17)²)

= √ (1 – 64/289)

= √ (289 – 64/289)

= √ (225/289)

= 15/17

Hence,

Cosec y = 1/sin y = 1/ (15/17) = 17/15

Therefore,

Cosec (cos^-1 8/17) = 17/15

(vi) Given Sec (sin^-1 12/13)

(vii) Given Tan (cos^-1 8/17)

(viii) Given cot (cos^-1 3/5)

(ix) Given Cos (tan^-1 24/7)

.

Chapter 4 Inverse Trigonometric Functions Ex 4.9

Q1. Evaluate:

(i) Cos {sin^-1 (-7/25)}

(ii) Sec {cot^-1 (-5/12)}

(iii) Cot {sec^-1 (-13/5)}

Solution:

(i) Given Cos {sin^-1 (-7/25)}

(ii) Given Sec {cot^-1 (-5/12)}

(iii) Given Cot {sec^-1 (-13/5)}

Chapter 4 Inverse Trigonometric Functions Ex 4.10

Q1. Evaluate:

(i) Cot (sin^-1 (3/4) + sec^-1 (4/3))

(ii) Sin (tan^-1 x + tan^-1 1/x) for x < 0

(iii) Sin (tan^-1 x + tan^-1 1/x) for x > 0

(iv) Cot (tan^-1 a + cot^-1 a)

(v) Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

Solution:

(i) Given Cot (sin^-1 (3/4) + sec^-1 (4/3))

(ii) Given Sin (tan^-1 x + tan^-1 1/x) for x < 0

(iii) Given Sin (tan^-1 x + tan^-1 1/x) for x > 0

(iv) Given Cot (tan^-1 a + cot^-1 a)

(v) Given Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

= 0

Q2. If cos^-1 x + cos^-1 y = π/4, find the value of sin^-1 x + sin^-1 y.

Solution:

Given cos^-1 x + cos^-1 y = π/4

Q3. If sin^-1 x + sin^-1 y = π/3 and cos^-1 x – cos^-1 y = π/6, find the values of x and y.

Solution:

Given sin^-1 x + sin^-1 y = π/3 ……. Equation (i)

And cos^-1 x – cos^-1 y = π/6 ……… Equation (ii)

Q4. If cot (cos^-1 3/5 + sin^-1 x) = 0, find the value of x.

Solution:

Given cot (cos^-1 3/5 + sin^-1 x) = 0

On rearranging we get,

(cos^-1 3/5 + sin^-1 x) = cot^-1 (0)

(Cos^-1 3/5 + sin^-1 x) = π/2

We know that cos^-1 x + sin^-1 x = π/2

Then sin^-1 x = π/2 – cos^-1 x

Substituting the above in (cos^-1 3/5 + sin^-1 x) = π/2 we get,

(Cos^-1 3/5 + π/2 – cos^-1 x) = π/2

Now on rearranging we get,

(Cos^-1 3/5 – cos^-1 x) = π/2 – π/2

(Cos^-1 3/5 – cos^-1 x) = 0

Therefore Cos^-1 3/5 = cos^-1 x

On comparing the above equation we get,

x = 3/5

Q5. If (sin^-1 x)² + (cos^-1 x)² = 17 π²/36, find x.

Solution:

Given (sin^-1 x)² + (cos^-1 x)² = 17 π²/36

We know that cos^-1 x + sin^-1 x = π/2

Then cos^-1 x = π/2 – sin^-1 x

Substituting this in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get

(sin^-1 x)² + (π/2 – sin^-1 x)² = 17 π²/36

Let y = sin^-1 x

y² + ((π/2) – y)² = 17 π²/36

y² + π²/4 – y² – 2y ((π/2) – y) = 17 π²/36

π²/4 – πy + 2 y² = 17 π²/36

On rearranging and simplifying, we get

2y² – πy + 2/9 π² = 0

18y² – 9 πy + 2 π² = 0

18y² – 12 πy + 3 πy + 2 π² = 0

6y (3y – 2π) + π (3y – 2π) = 0

Now, (3y – 2π) = 0 and (6y + π) = 0

Therefore y = 2π/3 and y = – π/6

Now substituting y = – π/6 in y = sin^-1 x we get

sin^-1 x = – π/6

x = sin (- π/6)

x = -1/2

Now substituting y = -2π/3 in y = sin^-1 x we get

x = sin (2π/3)

x = √3/2

Now substituting x = √3/2 in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get,

= π/3 + π/6

= π/2 which is not equal to 17 π²/36

So we have to neglect this root.

Now substituting x = -1/2 in (sin^-1 x)² + (cos^-1 x)² = 17 π²/36 we get,

= π²/36 + 4 π²/9

= 17 π²/36

Hence x = -1/2.

Chapter 4 Inverse Trigonometric Functions Ex 4.11

Q1. Prove the following results:

(i) Tan^-1 (1/7) + tan^-1 (1/13) = tan^-1 (2/9)

(ii) Sin^-1 (12/13) + cos^-1 (4/5) + tan^-1 (63/16) = π

(iii) tan^-1 (1/4) + tan^-1 (2/9) = Sin^-1 (1/ √5)

Solution:

(i) Given Tan^-1 (1/7) + tan^-1 (1/13) = tan^-1 (2/9)

Hence, proved.

(ii) Given Sin^-1 (12/13) + cos^-1 (4/5) + tan^-1 (63/16) = π

Consider LHS

Hence, proved.

(iii) Given tan^-1 (1/4) + tan^-1 (2/9) = Sin^-1 (1/ √5)

Q2. Find the value of tan^-1 (x/y) – tan^-1 {(x-y)/(x + y)}

Solution:

Given tan^-1 (x/y) – tan^-1 {(x-y)/(x + y)}

Chapter 4 Inverse Trigonometric Functions Ex 4.12

Q1. Evaluate: Cos (sin ^-1 3/5 + sin^-1 5/13)

Solution:

Given Cos (sin ^-1 3/5 + sin^-1 5/13)

We know that,

Chapter 4 Inverse Trigonometric Functions Ex 4.13

Q1. If cos^-1 (x/2) + cos^-1 (y/3) = α, then prove that 9x² – 12xy cos α + 4y² = 36 sin² α

Solution:

Given cos^-1 (x/2) + cos^-1 (y/3) = α

Hence, proved.

Q2. Solve the equation: cos^-1 (a/x) – cos^-1 (b/x) = cos^-1 (1/b) – cos^-1 (1/a)

Solution:

Given cos^-1 (a/x) – cos^-1 (b/x) = cos^-1 (1/b) – cos^-1 (1/a)

Chapter 4 Inverse Trigonometric Functions Ex 4.14

Q1. Evaluate the following:

(i) tan {2 tan^-1 (1/5) – π/4}

(ii) Tan {1/2 sin^-1 (3/4)}

(iii) Sin {1/2 cos^-1 (4/5)}

(iv) Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

Solution:

(i) Given tan {2 tan^-1 (1/5) – π/4}

(ii) Given tan {1/2 sin^-1 (3/4)}

(iii) Given sin {1/2 cos^-1 (4/5)}

(iv) Given Sin (2 tan ^-1 2/3) + cos (tan^-1 √3)

Q2. Prove the following results:

(i) 2 sin^-1 (3/5) = tan^-1 (24/7)

(ii) tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

(iii) tan^-1 (2/3) = ½ tan^-1 (12/5)

(iv) tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

(v) sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

(viii) 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

(ix) 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

(x) 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Solution:

(i) Given 2 sin^-1 (3/5) = tan^-1 (24/7)

Hence, proved.

(ii) Given tan^-1 ¼ + tan^-1 (2/9) = ½ cos^-1 (3/5) = ½ sin^-1 (4/5)

Hence, proved.

(iii) Given tan^-1 (2/3) = ½ tan^-1 (12/5)

Hence, proved.

(iv) Given tan^-1 (1/7) + 2 tan^-1 (1/3) = π/4

Hence, proved.

(v) Given sin^-1 (4/5) + 2 tan^-1 (1/3) = π/2

(vi) Given 2 sin^-1 (3/5) – tan^-1 (17/31) = π/4

(vii) Given 2 tan^-1 (1/5) + tan^-1 (1/8) = tan^-1 (4/7)

Hence, proved.

(viii) Given 2 tan^-1 (3/4) – tan^-1 (17/31) = π/4

Hence, proved.

(ix) Given 2 tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

Hence, proved.

(x) Given 4 tan^-1(1/5) – tan^-1(1/239) = π/4

Hence, proved.

Q3. If sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²), then prove that x = (a – b)/ (1 + a b)

Solution:

Given sin^-1 (2a/1 + a²) – cos^-1(1 – b²/1 + b²) = tan^-1(2x/1 – x²)

Hence, proved.

Q4. Prove that:

(i) tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

(ii) sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)} = 1

Solution:

(i) Given tan^-1{(1 – x²)/ 2x)} + cot^-1{(1 – x²)/ 2x)} = π/2

Hence, proved.

(ii) Given sin {tan^-1 (1 – x²)/ 2x) + cos^-1 (1 – x²)/ (1 + x²)}

Hence, proved.

Q5. If sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x, prove that x = (a + b/ 1 – a b)

Solution:

Given sin^-1 (2a/ 1+ a²) + sin^-1 (2b/ 1+ b²) = 2 tan^-1 x

Hence, proved.

Textbook Solutions are considered an extremely helpful resource for exam preparation. EduGrown.in gives its users access to a profuse supply of Textbook questions and their solutions. CBSE Class 6 Math Textbook Solutions are created by experts of the subject, hence, sure to prepare students to score well. The questions provided in Textbook Books are prepared in accordance with CBSE, thus holding higher chances of appearing on CBSE question papers. Not only do these Textbook Solutions for Class 6 Math strengthen students’ foundation in the subject, but also give them the ability to tackle different types of questions easily.

RD Sharma Solutions for class 6

• RD Sharma Solutions for Class 6 Maths Chapter 1 – Knowing Our Numbers
• RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing With Numbers
• RD Sharma Solutions for Class 6 Maths Chapter 3 – Whole Numbers
• RD Sharma Solutions for Class 6 Maths Chapter 4 – Operations on Whole Numbers
• RD Sharma Solutions for Class 6 Maths Chapter 5 – Negative Numbers and Integers
• RD Sharma Solutions for Class 6 Maths Chapter 6 – Fractions
• RD Sharma Solutions for Class 6 Maths Chapter 7 – Decimals
• RD Sharma Solutions for Class 6 Maths Chapter 8 – Introduction to Algebra
• RD Sharma Solutions for Class 6 Maths Chapter 9 – Ratio, Proportion and Unitary Method
• RD Sharma Solutions for Class 6 Maths Chapter 10 – Basic Geometrical Concepts
• RD Sharma Solutions for Class 6 Maths Chapter 11 – Angles
• RD Sharma Solutions for Class 6 Maths Chapter 12 – Triangles
• RD Sharma Solutions for Class 6 Maths Chapter 13 – Quadrilaterals
• RD Sharma Solutions for Class 6 Maths Chapter 14 – Circles
• RD Sharma Solutions for Class 6 Maths Chapter 15 – Pair of Lines and Transversal
• RD Sharma Solutions for Class 6 Maths Chapter 16 – Understanding Three Dimensional Shapes
• RD Sharma Solutions for Class 6 Maths Chapter 17 – Symmetry
• RD Sharma Solutions for Class 6 Maths Chapter 18 – Basic Geometrical Tools
• RD Sharma Solutions for Class 6 Maths Chapter 19 – Geometrical Constructions
• RD Sharma Solutions for Class 6 Maths Chapter 20 – Mensuration
• RD Sharma Solutions for Class 6 Maths Chapter 21 – Data Handling-I(Presentation of Data)
• RD Sharma Solutions for Class 6 Maths Chapter 22 – Data Handling-II(Pictographs)
• RD Sharma Solutions for Class 6 Maths Chapter 23 – Data Handling-III(Bar Graphs)

Textbook Solutions are considered an extremely helpful resource for exam preparation. EduGrown.in gives its users access to a profuse supply of Textbook questions and their solutions. CBSE Class 7 Math Textbook Solutions are created by experts of the subject, hence, sure to prepare students to score well. The questions provided in Textbook Books are prepared in accordance with CBSE, thus holding higher chances of appearing on CBSE question papers. Not only do these Textbook Solutions for Class 7 Math strengthen students’ foundation in the subject, but also give them the ability to tackle different types of questions easily.

RD Sharma Solutions for class 7

• RD Sharma Solutions for Class 7 Maths Chapter 1 – Integers
• RD Sharma Solutions for Class 7 Maths Chapter 2 – Fractions
• RD Sharma Solutions for Class 7 Maths Chapter 3 – Decimals
• RD Sharma Solutions for Class 7 Maths Chapter 4 – Rational Numbers
• RD Sharma Solutions for Class 7 Maths Chapter 5 – Operations on Rational Numbers
• RD Sharma Solutions for Class 7 Maths Chapter 6 – Exponents
• RD Sharma Solutions for Class 7 Maths Chapter 7 – Algebraic Expressions
• RD Sharma Solutions for Class 7 Maths Chapter 8 – Linear Equations in One Variable
• RD Sharma Solutions for Class 7 Maths Chapter 9 – Ratio and Proportion
• RD Sharma Solutions for Class 7 Maths Chapter 10 – Unitary Method
• RD Sharma Solutions for Class 7 Maths Chapter 11 – Percentage
• RD Sharma Solutions for Class 7 Maths Chapter 12 – Profit and Loss
• RD Sharma Solutions for Class 7 Maths Chapter 13 – Simple Interest
• RD Sharma Solutions for Class 7 Maths Chapter 14 – Lines and Angles
• RD Sharma Solutions for Class 7 Maths Chapter 15 – Properties of Triangles
• RD Sharma Solutions for Class 7 Maths Chapter 16 – Congruence
• RD Sharma Solutions for Class 7 Maths Chapter 17 – Constructions
• RD Sharma Solutions for Class 7 Maths Chapter 18 – Symmetry
• RD Sharma Solutions for Class 7 Maths Chapter 19 – Visualising Solid Shapes
• RD Sharma Solutions for Class 7 Maths Chapter 20 – Mensuration – I
• RD Sharma Solutions for Class 7 Maths Chapter 21 – Mensuration – II
• RD Sharma Solutions for Class 7 Maths Chapter 22 – Data Handling I (Collection and Organisation of Data)
• RD Sharma Solutions for Class 7 Maths Chapter 23 – Data Handling II (Central Values)
• RD Sharma Solutions for Class 7 Maths Chapter 24 – Data Handling III (Construction of Bar Graphs)
• RD Sharma Solutions for Class 7 Maths Chapter 25 – Data Handling IV (Probability)

Textbook Solutions are considered an extremely helpful resource for exam preparation. EduGrown.in gives its users access to a profuse supply of Textbook questions and their solutions. CBSE Class 8 Math Textbook Solutions are created by experts of the subject, hence, sure to prepare students to score well. The questions provided in Textbook Books are prepared in accordance with CBSE, thus holding higher chances of appearing on CBSE question papers. Not only do these Textbook Solutions for Class 8 Math strengthen students’ foundation in the subject, but also give them the ability to tackle different types of questions easily.

RD Sharma Solutions for class 8

• RD Sharma Solutions for Class 8 Maths Chapter 1 – Rational Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 2 – Powers
• RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots
• RD Sharma Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots
• RD Sharma Solutions for Class 8 Maths Chapter 5 – Playing with Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 6 – Algebraic Expressions and Identities
• RD Sharma Solutions for Class 8 Maths Chapter 7 – Factorization
• RD Sharma Solutions for Class 8 Maths Chapter 8 – Division of Algebraic Expressions
• RD Sharma Solutions for Class 8 Maths Chapter 9 – Linear Equation in One Variable
• RD Sharma Solutions for Class 8 Maths Chapter 10 – Direct and Inverse Variations
• RD Sharma Solutions for Class 8 Maths Chapter 11 – Time and Work
• RD Sharma Solutions for Class 8 Maths Chapter 12 – Percentage
• RD Sharma Solutions for Class 8 Maths Chapter 13 – Proft, Loss, Discount and Value Added Tax (VAT)
• RD Sharma Solutions for Class 8 Maths Chapter 14 – Compound Interest
• RD Sharma Solutions for Class 8 Maths Chapter 15 – Understanding Shapes-I (Polygons)
• RD Sharma Solutions for Class 8 Maths Chapter 16 – Understanding Shapes-II (Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 17 – Understanding Shapes-III (Special Types of Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 18 – Practical Geometry (Constructions)
• RD Sharma Solutions for Class 8 Maths Chapter 19 – Visualising Shapes
• RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon)
• RD Sharma Solutions for Class 8 Maths Chapter 21 – Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
• RD Sharma Solutions for Class 8 Maths Chapter 22 – Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
• RD Sharma Solutions for Class 8 Maths Chapter 23 – Data Handling-I (Classification and Tabulation of Data)
• RD Sharma Solutions for Class 8 Maths Chapter 24 – Data Handling-II (Graphical Representation of Data as Histograms)
• RD Sharma Solutions for Class 8 Maths Chapter 25 – Data Handling-III (Pictorial Representation of Data as Pie Charts or Circle Graphs)
• RD Sharma Solutions for Class 8 Maths Chapter 26 – Data Handling-IV (Probability)
• RD Sharma Solutions for Class 8 Maths Chapter 27 – Introduction to Graphs