RD Sharma Solutions for Class 12, maths Chapter 6

Class 12: Maths Chapter 6 solutions. Complete Class 12 Maths Chapter 6 Notes.

Contents

1 RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants
2 RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants: Download PDF
3 Chapterwise RD Sharma Solutions for Class 12 Maths :
4 About RD Sharma
- 4.1 Related

RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

Exercise 6.1 Page No: 6.10

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:

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants

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

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 30

= 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

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

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

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

Exercise 6.2 Page No: 6.57

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

= 2[1(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,

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

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.

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

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)

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

Solution:

Given,

Solution:

Given,

Solution:

Given,

Solution:

Given,

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

Solution:

Given,

Solution:

Given,

= a [a (a + x + y) + az] + 0 + 0

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

= 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.

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

Solution:

Given,

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

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.

Exercise 6.3 Page No: 6.71

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)

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

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)

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

⇒

⇒ 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

⇒ 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

⇒ – 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.

⇒ [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

Exercise 6.4 Page No: 6.84

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] + 1[2 – 4(– 4)]

⇒ D = 3[12 – 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 = 2[12 – 3] – 1[3 + 33] + 1[– 1 – 44]

⇒ D₁ = 2[9] – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 3[3 + 33] – 2[– 6 – 12] + 1[– 22 + 4]

⇒ D₂ = 3[36] – 2(– 18) – 18

⇒ D₂ = 126

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 3[44 + 1] – 1[– 22 + 4] + 2[2 + 16]

⇒ D₃ = 3[45] – 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] – 1[4 + 5(– 3)]

⇒ D = 1[– 5 – 4] + 4[8] – [– 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)] – 1[2 (39) – (– 5) (1)]

⇒ D₁ = 11[– 5 – 4] + 4[39 – 2] – 1[78 + 5]

⇒ D₁ = 11[– 9] + 4(37) – 83

⇒ D₁ = – 99 – 148 – 45

⇒ D₁ = – 34

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 1[39 – 2] – 11[2 + 6] – 1[2 + 117]

⇒ D₂ = 1[37] – 11(8) – 119

⇒ D₂ = – 170

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 1[– 5 – (39) (2)] – (– 4) [2 – (39) (– 3)] + 11[4 – (– 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] – 3[1 – 3(2)]

⇒ D = 6[12 + 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 = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

⇒ D₁ = 5[14] – 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 = 6[20 + 16] – 5[4 – 2(– 2)] + (– 3)[8 – 10]

⇒ D₂ = 6[36] – 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 = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

⇒ D₃ = 6[19] – 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 = 1[1] – 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 = 5[1] – 1[(3) (1) – (4) (1)] + 0[0 – (4) (1)]

⇒ D₁ = 5 – 1[3 – 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 = 1[3 – 4] – 5[– 1] + 0[0 – 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 = 1[4 – 0] – 1[0 – 3] + 5[0 – 1]

⇒ D₃ = 1[4] – 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 = 0[0] – 2[(0) (1) – 0] – 3[1 (4) – 3 (3)]

⇒ D = 0 – 0 – 3[4 – 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 = 0[0] – 2[(0) (– 4) – 0] – 3[4 (– 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 = 0[0] – 0[(0) (1) – 0] – 3[1 (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 = 0[9 – (– 4) 4] – 2[(3) (1) – (– 4) (3)] + 0[1 (4) – 3 (3)]

⇒ D₃ = 0[25] – 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)] + 1[2(6) – 3(– 8)]

⇒ D = 5[48 + 2] – 7[– 36 + 3] + 1[12 + 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 = 11[48 + 2] + 7[– 90 + 7] + 1[30 + 56]

⇒ D₁ = 11[50] + 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] + 1[42 – 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] + 11[12 + 24]

⇒ D₃ = 5[– 86] + 7[– 3] + 11[36]

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have

Exercise 6.5 Page No: 6.89

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,

RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants Image 263

= 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.

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions

RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants: Download PDF

RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants

Download PDF: RD Sharma Solutions for Class 12 Maths Chapter 6–Determinants PDF

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About RD Sharma

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

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

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RD Sharma Class 12 Solutions