# RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants

## RD Sharma Solutions for Class 12 Maths Chapter 6 – Get Free PDF Updated for 2023-24

RD Sharma Solutions for Class 12 Maths Chapter 6 – Determinants are designed by BYJU’S experts to boost confidence among students in understanding the concepts covered in the chapter. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants is provided here. RD Sharma Solutions are primarily designed for CBSE students and are based on the latest syllabus prescribed by the CBSE Board. Practising the textbook problems is an essential task to learn and score well in Mathematics.

Students are required to go through RD Sharma Solutions thoroughly before the final exams to score well and strengthen their problem-solving abilities. The RD Sharma Solutions for Class 12 of this chapter consists of five exercises and explains the concept of Determinants and their properties. Students can access these solutions in PDF format as per their requirements anytime.

## RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants:

### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 – Determinants

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:

Solution:

(i) Let Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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 Mij and Cij 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, Cij = (–1)i+j × Mij

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

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,

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.

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)

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

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 1st row

⇒ D = 5(1) – (– 3) (– 2)

⇒ D = 5 – 6

⇒ D = – 1

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 5(4) – (– 7) (– 2)

⇒ D1 = 20 – 14

⇒ D1 = 6

And

Solving determinant, expanding along 1st 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 1st row

⇒ D1 = 1(– 2) – (– 7) (– 1)

⇒ D1 = – 2 – 7

⇒ D1 = – 9

And

Solving determinant, expanding along 1st 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 1st row

⇒ D1 = 17(5) – (6) (– 1)

⇒ D1 = 85 + 6

⇒ D1 = 91

Solving determinant, expanding along 1st 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 1st row

⇒ D = 3(– 1) – (3) (1)

⇒ D = – 3 – 3

⇒ D = – 6

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 19(– 1) – (23) (1)

⇒ D1 = – 19 – 23

⇒ D1 = – 42

Solving determinant, expanding along 1st 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 1st 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 1st row

⇒ D = 3(a) – (2) (a)

⇒ D = 3a – 2a

⇒ D = a

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 4(a) – (2) (a)

⇒ D = 4a – 2a

⇒ D = 2a

Solving determinant, expanding along 1st 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 1st row

⇒ D = 2 (6) – (3) (1)

⇒ D = 12 – 3

⇒ D = 9

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 10 (6) – (3) (4)

⇒ D = 60 – 12

⇒ D = 48

Solving determinant, expanding along 1st 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 1st row

⇒ D = 5(6) – (7) (4)

⇒ D = 30 – 28

⇒ D = 2

Again,

Solving determinant, expanding along 1st row

⇒ D1 = – 2(6) – (7) (– 3)

⇒ D1 = – 12 + 21

⇒ D1 = 9

Solving determinant, expanding along 1st 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 1st row

⇒ D = 3(9) – (5) (– 2)

⇒ D = 27 + 10

⇒ D = 37

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 10(3) – (8) (5)

⇒ D1 = 30 – 40

⇒ D1 = – 10

Solving determinant, expanding along 1st 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 1st row

⇒ D = 1(1) – (3) (2)

⇒ D = 1 – 6

⇒ D = – 5

Again,

Solving determinant, expanding along 1st row

⇒ D1 = 1(1) – (2) (4)

⇒ D1 = 1 – 8

⇒ D1 = – 7

Solving determinant, expanding along 1st 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 1st 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 1st 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 1st row

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

⇒ D2 = 3[36] – 2(– 18) – 18

⇒ D2 = 126

Solving determinant, expanding along 1st 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 1st 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 1st 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 1st row

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

⇒ D2 = 1[37] – 11(8) – 119

⇒ D2 = – 170

And,

Solving determinant, expanding along 1st 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 1st 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 1st column by B matrices

Here

Solving determinant, expanding along 1st 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 1st 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 1st column by B matrices

Here

Solving determinant, expanding along 1st 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 Dj be the determinant obtained from D after replacing the jth 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 1st Row

⇒ D = 1[1] – 1[– 1] + 0[– 1]

⇒ D = 1 + 1 + 0

⇒ D = 2

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

Here

Solving determinant, expanding along 1st 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 1st 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 1st column by B matrices

Here

Solving determinant, expanding along 1st 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 1st 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 1st column by B matrices

Here

Solving determinant, expanding along 1st 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 2nd 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 3rd column by B matrices

Here

Solving determinant, expanding along 1st 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 1st 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 1st column by B matrices

Here

Solving determinant, expanding along 1st 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 2nd column by B matrices

Here

Solving determinant, expanding along 1st 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 3rd column by B matrices

Here

Solving determinant, expanding along 1st 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,

= 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.

### Also, Access RD Sharma Solutions for Class 12 Maths Chapter 6 Determinants

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

## RD Sharma Class 12 Solutions Chapter 6 Determinants

Let us have a look at some of the important concepts that are discussed in this chapter.

• Definition of determinants
• The determinant of a square matrix of orders 1, 2 and 3
• The determinant of a square matrix of order 3 by using the Sarrus diagram
• Definition and meaning of singular matrix
• Minors and cofactors of determinants
• Properties of determinants
• Evaluation of determinants
• Evaluation of determinants by using the properties of determinants and proving identities
• The solution of determinant equations
• Evaluation of determinants by using factor theorem
• Applications of determinants to coordinate geometry
• Area of triangle
• Conditions of collinearity of three points
• Equation of a line passing through two given points
• Applications of determinants in solving a system of linear equations
• The solution of a non-homogeneous system of linear equations
• Condition for consistency
• Solutions of a homogeneous system of linear equations

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Q1

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### Why is it important to download RD Sharma Solutions for Class 12 Maths Chapter 6 for students of the CBSE board?

RD Sharma Solutions are curated by a set of expert faculty at BYJU’S with the aim to provide the best study source for students. Professional teachers suggest students practise these solutions on a daily basis in order to strengthen their basics in the chapter Determinants. The questions are solved in a simple and concise manner to make learning fun and interesting for students. Following these solutions, students will be able to appear for other competitive exams with confidence.