NCERT Class 12 Mathematics Chapter 4 Determinants

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NCERT Class 12 Mathematics Chapter 4 Determinants

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Determinants

Chapter – 4

Exercise 4.1

Evaluate the determinants in Exercises 1 and 2. 

Ans: 

= 2(- 1) – 4(- 5) 

= – 2 + 20 

= 18

Ans: 

Ans:

Ans: 

Now,

Ans: 

The given matrix is It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C₁) for easier calculation.

Therefore 

5. Evaluate the determinants

Ans: Let

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

Hence,

Ans: Let

Hence,

Ans: Let

Hence,

Ans: Let

Hence,

= 2(0 – 5) – 0 – 3(1 + 4) 

= – 10 + 15 = 5

Ans: 

Let

Hence,

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

= 1(3 )- 1(- 3) – 2(3) 

= 3 + 3 – 6 = 0

7. Find values of x, if

Ans: 

Therefore,

⇒ 2 × 1 – 5 × 4 = 2x × x – 6 × 4

⇒ 2 – 20 = 2x² – 24

⇒ 2x² = 6 

⇒ x² = 3 

⇒ x = ± √3

:

Ans: Therefore,

⇒ 2 × 5 – 3 × 4 = x × 5 – 3 × 2x

⇒ 10 – 12 = 5x – 6x 

⇒ – 2 = – x

⇒ x = 2

(A) 6

(B) ± 6

(C) – 6

(D) 0

Ans: 

Therefore, 

⇒ x² – 36 = 36 – 36

⇒ x² – 36 = 0

⇒ x² = 36

⇒ x = ± 6

Thus, the correct option is B.

EXERCISE 4.2

1. Find area of the triangle with vertices at the point given in each of the following: 

(i) (1,0), (6,0), (4,3)

Ans: The area of the triangle with vertices (1,0),(6,0), (4,3) is given by the relation,

= 1/2 [1(0 – 3) – 0(6 – 4) + 1(18 – 0)]

= 1/2 [- 3 + 18]

= 1/2 [15] 

= 15/2

Hence, area of the triangle is 15/2 square units.

(ⅱ) (2,7), (1, 1), (10,8) 

Ans: The area of the triangle with vertices (2,7), (1,1), (10,8) is given by the relation,

= 1/2 [2(1 – 8) – 7(1 – 10) + 1(8 – 10)] 

= 1/2 [2(- 7) – 7(- 9) + 1(- 2)]

= 1/2 [- 14 + 63 – 2] 

= 12 [47] 

= 47/2

Hence, area of the triangle is 47/2 square units.

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

Ans: The area of the triangle with vertices (- 2,- 3),(3,2), (- 1, – 8) is given by the relation,

= 1/2 [- 2(2 + 8) + 3(3 + 1) + 1(- 24 + 2)] 

= 1/2[- 2(10) + 3(4) + 1(- 22)] 

= 1/2 [- 20 + 12 – 22] 

= – 1/2 [30] 

= – 15

Hence, area of the triangle is 15 square units.

2. Show that points:

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Ans: The area of the triangle with vertices A(a,b + c), B(b, c + a), C(c, a + b) is given by the absolute value of the relation:

Thus, the area of the triangle formed by points is zero. 

Hence, the points are collinear.

3. Find values of k if area of triangle is 4 square. units and vertices are:

(i) (k, 0), (4, 0), (0, 2)

(ii) (- 2, 0), (0,4), (0, k)

Ans: We know that the area of a triangle whose vertices are (x₁,y₁) (x₂ ,y₂) and (x₃, y₃) is the absolute value of the determinant (∆), where 

It is given that the area of triangle is 4 square units.

Hence, = ± 4

(i) The area of the triangle with vertices (k.0),(4,0), (0,2) is given by the relation,

Therefore, – k + 4 = ± 4 

When – k + 4 = – 4 

Then k = 8

When – k + 4 = – 4 

Then k = 0 

Hence, k = 0,8

(ii) (- 2, 0), (0,4), (0, k)

Ans: The area of the triangle with vertices (- 2,0), (0,4), (0,k) is given by the relation,

= k – 4 

Therefore, – k + 4 = ± 4 

When k – 4 = 4 

Then k = 8 

When k – 4 = – 4 

Then k = 0 

Hence, k = 0,8

4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.

Ans: Let P(x,y) be any point on the line joining points A (1,2) and B(3,6). Then, the points A, B and P are collinear. 

Hence, the area of triangle ABP will be zero. 

Therefore, 

⇒ 1/2 [1(6 – y) – 2(3 – x) + 1(3y – 6x)] = 0 

⇒ 6 – y – 6 + 2x + 3y – 6x = 0 

⇒ 2y – 4x = 0 

⇒ y = 2x 

Thus, the equation of the line joining the given points is y = 2x. 

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Ans: Let P(x,y) be any point on the line joining points A (3,1) and B(9.3). Then, the points A, B and P are collinear. Hence, the area of triangle ABP will be zero. Therefore,

⇒ 1/2[3(3 – y) – 1(9 – x) + 1(9y – 3x)] = 0 

⇒ 9 – 3y – 9 + x + 9y – 3x = 0

⇒ 6y – 2x = 0 

⇒ x – 3y = 0 

Thus, the equation of the line joining the given points is x – 3y = 0

5. If area of triangle is 35 sq units with vertices (2,- 6), (5, 4) and (k, 4). Then k is 

(A) 12 

(B) – 2 

(C) – 12, – 2 

(D) 12, – 2

Ans: The area of the triangle with vertices (2,- 6), (5.4), (k,4) is given by the relation,

= 1/2[2(4 – 4) + 6(5 – k) + 1(20 – 4k)] 

= 1/2[30 – 6k + 20 – 4k] 

= 1/2[50 – 10k]

=  25 – 5k 

It is given that the area of the triangle is 35 square units 

Hence, ∆ = ± 35

Therefore

⇒ 25 – 5k = ± 35

⇒ 5(5 – k)= ± 35 

⇒ 5 – k = ± 7 

When, 5 – k = – 7 

Then, k = 12 

When, 5 – k = 7 

Then, k = – 2 

Hence, k = 12, – 2 

Thus, the correct option is D.

EXERCISE 4.3

Write Minors and Cofactors of the elements of following determinants:

Ans: (i) The given determinant is 

M₁₁ = minor of element a₁₁ = 3 

M₁₂ = minor of element a₁₂ = 0

M₂₁ = minor of element a₂₁ = – 4

M₂₂ =  minor of element a₂₂ = 2

Ans: 

M₁₁ =  minor of element a₁₁ = d 

M₁₂  = minor of element a₁₂ = b 

M₂₁ = minor of element a₂₁ = c

M₂₂ = minor of element a₂₂ = a

Ans: 

Ans: 

Ans: 

We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors. 

Therefore,

∆ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃

= 2(7) + 0(7) + 1( – 7) 

= 14 – 7 

= 7

Ans: 

We know that ∆ is equal to the sum of the product of the elements of the third column with their corresponding cofactors. 

Therefore,

Hence,

∆ = (x – y)(y – z)(z − x) 

Ans: We know that ∆ is equal to the sum of the product of the elements of a column or row with their corresponding cofactors. 

∆ = a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁

Thus, the correct option is D.

EXERCISE 4.4

Find adjoint of each of the matrices in Exercises 1 and 2.

1.

Ans: 

Ans: 

Therefore,

3. 

Ans: 

Ans: 

Now,

Also,

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

5.

Ans: 

Then,

Ans: 

Hence,

7.

Ans: 

Ans: 

Ans: 

Now, 

Ans: 

Then, expanding along C₁,

|A| = 1(8 – 6) – 0 + 3(3 – 4) = 2 – 3 

= – 1

Now, 

Therefore,

Hence, 

Ans:

Now,

Therefore,

Hence, 

Ans: 

Now, 

Also,

Hence, proved.

Ans:

Hence, A² – 5A + 7 I = 0

Now,

Ans: 

Hence,

Comparing the corresponding elements of the two matrices, we have:

⇒ -1/b h = – 1

⇒ b = 1

Also,

 ⇒ – 3 – a/b = 1

⇒ – 3 – a = 1 

⇒ a = – 4 

Thus, a = – 4 and b = 1

 Show that A³ – 6A² + 5A + 11 I = O. Hence, find A⁻¹.

Ans:

Therefore,

And,

Thus, A³ – 6A² + 5A + 11I = 0 

Now, 

⇒ A³ – 6A² + 5A + 11I = 0 

⇒ (AAA)A⁻¹ – 6(AA)A⁻¹ + 5AA⁻¹ + 11IA⁻¹ = 0      [post – multiplying by A⁻¹ as |A| ≠ 0]

⇒ ΑΑ(ΑΑ⁻¹) – 6Α(AA⁻¹) + 5(AA⁻¹) = – (IA⁻¹) 

⇒ A² – 6A + 5I – 11A⁻¹

⇒ A⁻¹ = – 1/11(A² – 6A + 5I) ……(1)

Now,

From equation (1) and (2)

Verify that A³ – 6A² + 9A + 4 I = O and Hence, find A⁻¹

Ans: 

Therefore,

A² = A.A

And

A³ = A².A

Now,

Thus,

 A³ – 6A² + 9A – 4 I = 0

Now, 

Now,

From equations (1) and (2),

17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to

(A) |A|

(B) |A|²

(C) |A|³

(D) 3 |A|

Ans: A be a non-singular square matrix of order 3 × 3

(adjA) A = |A| I

Therefore, 

|adjA| = |A|² 

Thus, the correct option is B.

18. If A is an invertible matrix of order 2, then det (A⁻¹) is equal to 

(A) det (A) 

(B) 1/det (A) 

(C) 1 

(D) 0

Ans: 

Then,

|A| = ad – bc 

And, 

Now,

Hence, 

Hence,

det(A⁻¹) = 1/det (A)

Thus, the correct option is B.

EXERCISE 4.5

Examine the consistency of the system of equations in 1 to 6

1. x + 2y = 2 

2x + 3y = 3

Ans: The given system of equations is: 

x + 2y = 2 

2x + 3y = 3

The given system of equations can be written in the form of AX = B, where

Hence, 

|A| = 1(3) – 2(2) 

=  3 – 4 

= – 1

≠ 0

So, A is non-singular. 

Therefore, A⁻¹  exists. 

Thus, the given system of equations is consistent.

2. 2x – y = 5

x + y = 4

Ans: The given system of equations is: 

2x – y = 5

x + y = 4

The given system of equations can be written in the form of AX = B, where

Hence,

|A| = 2(1) – 1(- 1)

= 2 + 1

= 3 

≠ 0 

So, A is non-singular. 

Therefore, A⁻¹ exists. 

Hence, the given system of equations is consistent.

3. x + 3y = 5 

2x + 6y = 8

Ans: The given system of equations is: 

x + 3y = 5

2x + 6y = 8

The given system of equations can be written in the form of AX = B, where

Hence,

|A| = 1(6) – 3(2) 

= 6 – 6 

= 0 

So, A is a singular matrix. 

Now, 

Therefore,

Thus, the solution of the given system of equations does not exist. 

Hence, the system of equations is inconsistent.

4. x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

Ans: The given system of equations is:

 x + y + z = 1

 2x + 3y + 2z = 2 

ax + ay + 2az = 4

The given system of equations can be written in the form of AX = B, where

Hence, |A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) 

= 4a – 2a – a

= 4a – 3a 

= a ≠ 0 

So, A is non-singular.

Therefore, A⁻¹ exists.

Thus, the given system of equations is consistent.

5. 3x – y – 2z = 2

2y – z = – 1 

3x – 5y = 3 

Ans: The given system of equations is:

 3x – y – 2z = 2

2y – z = – 1 

3x – 5y = 3 

The given system of equations can be written in the form of AX = B, where

Hence, 

|A| = 3(0 – 5) – 0 + 3(1 + 4) 

= – 15 + 15

= 0 

So, A is a singular matrix.

Now,

Therefore, 

Thus, the solution of the given system of equations does not exist. 

Hence, the system of equations is inconsistent.

6. 5x – y + 4z = 5

2x + 3y + 5z = 2 

5x – 2y + 6z = – 1

Ans: The given system of equations is: 

5x – y + 4z = 5

2x + 3y + 5z = 2 

5x – 2y + 6z = – 1

The given system of equations can be written in the form of AX = B, where

Hence, |A| = 5(18 + 10) + 1(12 – 25) + 4(- 4 – 15) 

= 5(28) + 1(- 13) + 4(- 19) 

= 140 – 13 – 76

= 51 ≠ 0 

So, A is nonsingular.

Therefore, A⁻¹ exists. 

Hence, the given system of equations is consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14. 

7. 5x + 2y = 4 

7x + 3y = 5

Ans: The given system of equations is: 

5x + 2y = 4 

7x + 3y = 5

The given system of equations can be written in the form of AX = B, where

Hence, |A| = 15 – 14 

= 1 

≠ 0 

So, A is non-singular. 

Therefore, A⁻¹ exists. 

Now,

Then,

Hence, x = 2 and y = – 3

8. 2x – y = – 2 

3x + 4y = 3

Ans: The given system of equations is: 

 2x – y = – 2 

3x + 4y = 3 

The given system of equations can be written in the form of AX = B, where

Hence,

|A| = 8 + 3 

= 11 

≠ 0 

So, A is non-singular.

Therefore, A⁻¹ exists. 

Now,

Therefore,

Hence, x = – 5/11 and y = 12/11

9. 4x – 3y = 3 

3x + 5y = 7

Ans: The given system of equations is: 

4x – 3y = 3

3x – 5y = 7 

The given system of equations can be written in the form of AX = B, where

Hence,

|A| = – 20 + 9

= – 11 

≠ 0

So, A is nonsingular.

Therefore, A⁻¹ exists. 

Now,

Therefore, 

Hence, x = – 6/11 and y = -19/11

10. 5x + 2y = 3 

3x + 2y = 5

Ans: The given system of equations is: 

5x + 2y = 3 

3x + 2y = 5

The given system of equations can be written in the form of AX = B, where,

Hence,

|A| = 10 – 6 

= 4 

≠ 0

So, A is non-singular. 

Therefore, A⁻¹ exists.

Now, 

Therefore, 

Hence, x = – 1 and y = 4

11. 2x + y + z = 1

 x – 2y – z = 3/2

3y – 5z = 9

Ans: The given system of equations is: 

2x + y + z = 1

 x – 2y – z = 3/2

3y – 5z = 9

The given system of equations can be written in the form of AX = B, where

Hence, 

|A| = 2(10 + 3) – 1(- 5 – 3) + 0

= 2(13) – 1(- 8) 

= 26 + 8 

= 34 

≠ 0 

So, A is non-singular. 

Therefore, A⁻¹ exists. 

Now,

Hence,

Therefore,

= X = A⁻¹ B

Hence, x = 1, y = 1/2 and z = – 3/2

12. x – y + z = 4

2x + y – 3z = 0

x + y + z = 2

Ans: The given system of equations is: 

x – y + z = 4

2x + y – 3z = 0

x + y + z = 2

The given system of equations can be written in the form of AX = B, where

Hence, 

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) 

= 4 + 5 + 1

= 10 

≠ 0 

So, A is nonsingular.

Therefore, A⁻¹ exists.

Now,

Hence, 

Therefore,

= X = A⁻¹ B

Hence, x = 2, y = – 1 and z = 1

13. 2x + 3y + 3 = 5

x – 2y + z = – 4 

3x – y – 2z = 3

Ans: The given system of equations is:

2x + 3y + 3z = 5

x – 2y + z = – 4 

3x – y – 2z = 3

The given system of equations can be written in the form of AX = B, where

Hence, 

|A| = 2(4 + 1) – 3(- 2 – 3) + 3(- 1 + 6) 

= 10 + 15 + 5 

= 40 

≠ 0 

So, A is non-singular.

Therefore, A⁻¹ exists.

Now,

Hence,

Therefore, 

= X = A⁻¹ B

Hence, x = 1,y = 2 and z = – 1

14. x – y + 2z = 7 

3x + 4y – 5z = – 5 

2x – y + 3z = 12

Ans: The given system of equations is: 

x – y + 2z = 7 

3x + 4y – 5z = – 5 

2x – y + 3z = 12

The given system of equations can be written in the form of AX = B, where

Hence, 

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

= 7 + 19 – 22 

= 4

≠ 0 

So, A is non-singular.

Therefore, A⁻¹ exists.

Now,

Hence, 

Therefore, 

= X = A⁻¹ B

Hence, x = 2,y = 1 and z = 3

find A⁻¹ Using A⁻¹ solve the system of equations 

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

Ans: 2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

It is given that 

Therefore, 

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

= 0 – 6 + 5 

= – 1

≠ 0 

Now,

Hence,

The given system of equations can be written in the form of AX =  B, where

The solution of the system of equations is given by X = A⁻¹ B. 

Therefore,

= X = A⁻¹ B

Hence, x = 1, y = 2 and z = 3

16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method. 

Ans: Let the cost of onions, wheat, and rice per kg in ₹ be x, y and z respectively. 

Then, the given situation can be represented by a system of equations as: 

4x + 3y + 2z = 60

2x + 4y + 6z = 90 

6x + 2y + 3z = 70

The given system of equations can be written in the form of AX = B, where

Therefore, 

|A| = 4(12 – 12) – 3(6 – 36) + 2(4 – 24) 

= 0 + 90 – 40

= 50

≠ 0

So, A is non-singular.

Therefore, A⁻¹ exists. 

Now,

Therefore, 

Hence,

= X = A⁻¹ B

Thus, x = 5, y = 8 and z = 8

Hence, the cost of onions is ₹ 5 per kg the cost of wheat is ₹ 8 per kg, and the cost of rice is ₹ 8 per kg.

Miscellaneous Exercises on Chapter- 4

1.

Ans: 

= x(-x² – 1) – sin θ (- x sin  θ – cos θ) + cos θ (- sin θ + x cos θ) 

= – x³ – x + x sin² θ + sinθ cosθ – sine θ cos θ + x cos² θ

= – x³ – x + x(sin²θ + cos²θ)

= – x³ – x + x

= – x³

Hence, ∆ is independent of θ.

2. Evaluate

Ans: 

Let

Expanding along C₃,

3.

Ans: We know that (AB)⁻¹ = B⁻¹ A⁻¹.

It is given that Therefore, 

|B| = 1(3) – 2(- 1) – 2(- 2) 

= 3 + 2 – 4 

= 5 – 4 = 1 

Now,

Hence,

Now, 

Therefore,

(AB)⁻¹ = B⁻¹ A⁻¹

Thus, 

4. Let A verify that

(i) [adjA]⁻¹ = adj (A)⁻¹

(ii) (A⁻¹)⁻¹ = A

Ans: 

It is given that

Therefore,

|A| = 1(15 – 1) + 2(- 10 – 1) + 1(- 2 – 3) 

= 14 – 22 – 5

= – 13

Now,

Hence,

Now, 

(i) |adjA| = 14(- 4 – 9) – 11(- 11 – 15) – 5(- 33 + 20)

 = 14(- 13) – 11(- 26) – 5(- 13) 

= – 182 + 286 + 65

= 169 

We have,

Therefore, 

Now, 

Therefore, 

(ii)

Hence,

Now,

Therefore, 

Ans: 

Ans:

Using properties of determinants in Exercises 11 to 15, prove that:

7. Solve the system of equations

Choose the correct answer in Exercise 17 to 19.

Ans:

Then the given system of equations is as follows:

2p + 3q + 10r = 4

4p – 6q + 5r = 1 

6p + 9q – 20r = 2 

This system can be written in the form of AX = B, where

Therefore,

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

= 150 + 330 + 720 

= 1200 

Thus, A is non-singular. 

Therefore, A⁻¹ exists. 

Now,

Hence,

Now,

Therefore,

P =  1/2, q = 1/3 and r = 1/5

Hence, x = 2, y = 3 and z = 5

8. If x, y, z are nonzero real numbers, than the inverse of matrix

Ans:

It is given that Hence,

|A| = x(yz – 0) 

= xyz

 ≠ 0 

Now,

Therefore, 

(A) Det(A) = 0 

(B) Det(A) € (2,0) 

(C) Det(A) € (2,4)

(D) Det(A) [2,4]

Ans: 

It is given that Hence,

Therefore,

Det (A) € (2,4)

Thus, the correct option is D.