Question

NCERT Solutions Class 12 Maths Chapter 4 Determinants is designed by the experts of the subject matter who have years of experience in their field. Our NCERT Solutions is one of the best study materials one must refer to while preparing for their board and competitive exams. NCERT Solutions Class 12 is made with utter consideration to making it easy for students to go through all topics.

• Determinants – such scalar values which can be used to calculate the determinant with the use of the element of a square matrix. A square matrix is represented in the form of

\(\begin{vmatrix} a & b\\ c & d \end{vmatrix} \)

In the case of a \(3\times 3\) a square matrix, the elements are arranged in the form of

\(\begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)

• Properties of Determinants – There are several properties of the determinant which determine the methods of calculation which can be used to calculate the determinant.
  1. When the row and column of the determinant are interchanged with each other then there is no change in the value of the determinant.
  2. When the two rows and columns interchanged in a determinant then the sign of the determinant also changes.
  3. The value of the determinant is 0 when any two rows of columns are interchanged.
  4. The value of the determinant is increased by a multiple of K when any row or the column is multiplied by another variable K.
  5.  A determinant can be expressed as the sum of two or more two determinants if some or all elements of a row or column are expressed in the form of a sum of two or more terms.
• Applications of Determinants and Matrices – the main use of application of determinant of the matrix in checking the consistency of the system of linear equations.
  1. If a system of equations has no solutions then the system of equations is said to be inconsistent.
  2.  If a system has one or more than one solution then the system of equation is called a consistent equation.

NCERT Solutions Class 12 Maths is made in a very concise manner for easy and fast learning and revision.

Answer 1

NCERT Solutions Class 12 Maths Chapter 4 Determinants

1. Evaluate the determinant.

Answer:

Expanding along R₁, we get

= 2 x (-1) - 4 x (-5) 

= -2 + 20

 = 18

2. Evaluate the determinant.

Answer:

(i)

(ii)

3. If 

,

then show that |2A| = 4|A|

Answer:

Expanding along R₁, we get

= 2 x 4 - 4 x 8 

= 8 - 32

= -24        ......(1)

Expanding along R₁, we get

= 4(1 x 2 - 2 x 4)

= 4(--6)

= -24     ......(2)

From the equation (1) and (2), we get, 

|2A| = 4|A|

4. If 

,

then show that |3A| = 27|A|

Answer:

Expanding along R₁, we get

= 3(36 - 0) - 0(0-0) + 1(0 - 0) 

 = 108    ......(1)

Expanding along R₁, we get

= 27{1(4 - 0) - 0(0 - 0) + 1( 0 - 0)}

= 27(4)

= 108    ......(2)

From the equation (1) and (2), we get,

|3A| = 27|A|

5. Evaluate the determinants.

Answer:

Expanding along R₁, we get

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

= -15  + 3 - 0

= -12

Expanding along R₁, we get

= 3(1 + 6) + 4 (1 + 4) - 5(3 - 2)

= 21 + 20  + 5

 = 46

Expanding along R₁, we get

= 0(0 + 9) - 1(0 - 6) + 2(-3 -0)

= 0 + 6 - 6

= 0

Expanding along R₁, we get

= 2(0 - 5) + 1(0 + 3) - 2(0 - 6) 

= - 10 + 3 + 12

= 5

6. If

,

find |A|.

Answer:

Expanding along R₁, we get

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

= 3 + 3 - 6

= 0

7. Find the value of x, if

Answer:

⇒ 2 - 20 = 2x² - 24

⇒ x² = 3

⇒ x = \(\pm \sqrt3\)

⇒ 10 -12 = 5x - 6x

⇒ -2 = -x

⇒ x = 2

8. If

,

the x is equal to:

(A) 6

(B) \(\pm\) 6

(C) - 6

(D) 0

Answer:

⇒ x² - 36 = 36 - 36

⇒ x² = 36

⇒ x = \(\pm\) 6

Hence, option (B) is correct.

9. Using the property of determinants and without expanding, prove that:

Answer:

10. Using the property of determinants and without expanding, prove that:

Answer:

Answer 2

11. Using the property of determinants and without expanding, prove that:

Answer:

12. Using the property of determinants and without expanding, prove that:

Answer:

13. Using the property of determinants and without expanding, prove that:

Answer:

14. By using the properties of determinants, show that:

Answer:

15. By using the properties of determinants, show that:

Answer:

16. By using the properties of determinants, show that:

Answer:

17. By using the properties of determinants, show that:

Answer:

18. By using the properties of determinants, show that:

Answer:

19. By using the properties of determinants, show that:

Answer:

20. By using the properties of determinants, show that:

Answer:

21. By using the properties of determinants, show that:

Answer:

22. By using the properties of determinants, show that:

Answer:

23. By using the properties of determinants, show that:

Answer:

24. By using the properties of determinants, show that:

Answer:

25. By using the properties of determinants, show that:

Answer:

26. Let A be a square matrix of order 3 x 3, then |kA| is equal to:

(A) k|A|

(B) k²|A|

(C) k³|A|

(D) 3k|A|

Answer:

If B be a square matrix of order n x n, then |kB| = k^n-1|B|

Therefore, k|A|= k^3-1|A| = k²|A|

Hence, the option (B) is correct.

27. Which is the following is correct

(A) Determinant is a square matrix.

(B) Determinant is a number associated to a matrix.

(C) Determinant is a number associated to a square matrix.

(D) None of these

Answer:

Determinant is a number associated to a square matrix.

Hence, the option (C) is correct.

Answer 3

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

(i) (1, 0), (6, 0), (4, 3) 

(ii) (2, 7), (1, 1), (10, 8) 

(iii) (−2, −3), (3, 2), (−1, −8)

Answer:

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

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

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8) is given by the relation,

Hence, the area of the triangle is |-15| = 15 square units

29. Show that points are A(a, b + c), B(b, c + a), C(c, a + b) collinear.

Answer:

Area of ΔABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero. 

Hence, the points A, B, and C are collinear.

30. 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)

Answer:

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 (Δ), where

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

∴ Δ = ± 4. 

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

∴ −k + 4 = ± 4 

When −k + 4 = − 4, k = 8. 

When −k + 4 = 4, k = 0. 

Hence, k = 0, 8. 

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

∴ k − 4 = ± 4 

When k − 4 = − 4, k = 0. 

When k − 4 = 4, k = 8. 

Hence, k = 0, 8.

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

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

Answer:

(i) 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. Therefore, the area of triangle ABP will be zero.

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

(ii) 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. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is x − 3y = 0.

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

A. 12 

B. −2 

C. −12, −2 

D. 12, −2

Answer:

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

It is given that the area of the triangle is ±35. 

Therefore, we have:

⇒ 25 - 5k = \(\pm\) 35

⇒ 5(5 - k) = \(\pm\) 35

⇒ 5 - k = \(\pm\) 7

When 5 − k = −7, k = 5 + 7 = 12. 

When 5 − k = 7, k = 5 − 7 = −2. 

Hence, k = 12, −2. 

The correct answer is D.

33. Write Minors and Cofactors of the elements of following determinants:

(i) 

(ii)

Answer:

(i) The given determinant is 

Minor of element a_ij is M_ij. 

∴ M₁₁ = minor of element a₁₁ = 3 

M₁₂ = minor of element a₁₂ = 0 

M₂₁ = minor of element a₂₁ = −4 

M₂₂ = minor of element a₂₂ = 2 

Cofactor of a_ij is A_ij = (−1)^{i + j} M_ij. 

∴ A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² (3) = 3

A₁₂ = (−1)¹⁺² M₁₂ = (−1)³ (0) = 0 

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ (−4) = 4 

A₂₂ = (−1)²⁺² M₂₂ = (−1)⁴ (2) = 2

(ii) The given determinant is 

. 

Minor of element a_ij is M_ij. 

∴ M₁₁ = minor of element a₁₁= d 

M₁₂ = minor of element a₁₂= b 

M₂₁ = minor of element a₂₁= c 

M₂₂ = minor of element a₂₂= a 

Cofactor of a_ij is A_ij = (−1)^{i + j} M_ij. 

∴ A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² (d) = d 

A₁₂ = (−1)¹⁺² M₁₂ = (−1)³ (b) = −b 

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ (c) = −c 

A₂₂ = (−1)²⁺² M₂₂ = (−1)⁴ (a) = a

Answer 4

34. Write Minors and Cofactors of the elements of following determinants:

Answer:

Here,

and A_ij = (-1)^{i + j} M_ij, therefore

 

Here,

and A_ij = (-1)^{i + j} M_ij, therefore

35. Using Cofactors of elements of second row, evaluate 

Answer:

= a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

Here, a₂₁ = 2, a₂₂ = 0, a₂₃ = 1 and

Therefore,

= 2(7) + 0(7) + 1(-7) 

= 7

36. Using Cofactors of elements of third row, evaluate 

Answer:

= a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

Here, a₁₃ = yz, a₂₃ = zx, a₃₃ = xy and

Therefore,

= yz(z - y) + zx(x - z) + xy(y - x)

37. If 

A_ij is cofactors of a_ij, then value of △ is given by 

(A) a₁₁A₁₁ + a₁₂A₃₂ + a₁₃A₃₃

(B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁

(C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃

(D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Answer:

We know that:

Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors

∴Δ = a₁₁A₁₁ + a₂₁A₂₁ + _a31A₃₁ 

Hence, the value of Δ is given by the expression given in alternative D. 

The correct answer is D.

38. Find adjoint of each of the matrices.

Answer:

Let A = 

We have,

39. Find adjoint of each of the matrices.

Answer:

Let A = 

We have,

Hence,

40. Verify A (adj A) = (adj A) A = |A| I .

Answer:

Let  A = 

We have,

Now,

Now,

Hence, A(adj A) = (adj A) A = |A| I.

41. Verify A (adj A) = (adj A) A = |A| I .

Answer:

Now,

Now,

Also,

Hence, A(adj A) = (adj A) A = |A| I.

42. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

|A| = 6 + 8 = 14

Now,

43. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

|A| = -2 + 15 = 13

Now,

Answer 5

44. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) =10

Now,

45. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

|A| = 1(-3 -0) -0 + 0 = -3

Now,

46. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

Now,

47. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

By expanding along C₁, we have:

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

Now,

48. Find the inverse of each of the matrices (if it exists).

Answer:

Let  A =

We have,

|A| = 1(-cos²a - sin²a) = -(cos²a + sin²a) = -1

Now,

49. Let

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

Answer:

Let A =

We have,

|A| = 15 - 14 = 1

Now,

Now, let B =

We have,

|B| = 54 - 56 = -2

Now,

Then,

Therefore, we have |AB| = 67 x 61 - 87 x 47 = 4087 - 4089 = -2.

Also,

From (1) and (2), we have: 

(AB)⁻¹ = B⁻¹A⁻¹ 

Hence, the given result is proved.

50. If 

,

show that A²- 5A + 7I. Hence find A^-1.

Answer:

Hence,

51. For the matrix 

, 

find the numbers a and b such that A2 + aA + bI = O.

Answer:

Now,

Now,

We have:

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

\(- \frac1b= -1 \)

⇒ b =1

\(\frac{-3-a}{b}=1\)

⇒ - 3 - a = 1

⇒ a = -4

Hence, −4 and 1 are the required values of a and b respectively.

52. For the matrix 

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

Answer:

Thus, A³ - 6A² + 5A + 11I = O.

Now,

Now,

A² - 6A + 5I

From equation (1), we have:

Answer 6

53. If 

verify that A³ − 6A² + 9A − 4I = O and hence find A⁻¹

Answer:

Now,

A³ − 6A² + 9A − 4I

Now,

From equation (1), we have:

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

A. |A|

B. |A|²

C. |A|³

D. 3|A|

Answer:

We know that,

Hence, the correct answer is B.

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

A. det (A) 

B. \(\frac1 {det(A)}\)

C. 1 

D. 0

Answer:

Since A is an invertible matrix, A^-1 exists and A^-1 = \(\frac1{|A|}adjA.\)

As matrix A is order 2, let 

Then |A| = ad - bc and 

Now,

Hence, the correct answer is B.

56. Examine the consistency of the system of equations. 

x + 2y = 2 

2x + 3y = 3

Answer:

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

Now,

|A| = 1(3) - 2(2) = 3 - 4 = - 1 \(\ne\) 0

∴ A is non-singular. 

Therefore, A⁻¹ exists. 

Hence, the given system of equations is consistent.

57. Examine the consistency of the system of equations. 

2x − y = 5 

x + y = 4

Answer:

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

Now,

|A| = 2(1) - (-1)(1) = 2 + 1 = 3 \(\ne\) 0

∴ A is non-singular. 

Therefore, A−1 exists. 

Hence, the given system of equations is consistent.

58. Examine the consistency of the system of equations. 

x + 3y = 5 

2x + 6y = 8

Answer:

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

Now,

|A| = 1(6) - 3(2) = 6 - 6 = 0

∴ A is a singular matrix.

Now,

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

59. Examine the consistency of the system of equations. 

x + y + z = 1 

2x + 3y + 2z = 2 

ax + ay + 2az = 4

Answer:

The given system of equations is: 

x + y + z = 1 

2x + 3y + 2z = 2 

ax + ay + 2az = 4 

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

Now,

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

= 4a - 2a - a = 4a - 3a = a \(\ne\) 0

∴ A is non-singular.

Therefore, A−1 exists. 

Hence, the given system of equations is consistent.

60. Examine the consistency of the system of equations. 

3x − y − 2z = 2 

2y − z = −1 

3x − 5y = 3

Answer:

The given system of equations is: 

3x − y − 2z = 2 

2y − z = −1 

3x − 5y = 3 

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

Now,

|A| = 3(0 - 5) - 0 + 3(1 + 4) = - 15 + 15 = 0

∴ A is a singular matrix.

Now,

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

61. Examine the consistency of the system of equations. 

5x − y + 4z = 5 

2x + 3y + 5z = 2 

5x − 2y + 6z = −1

Answer:

The given system of equations is: 

5x − y + 4z = 5 

2x + 3y + 5z = 2 

5x − 2y + 6z = −1 

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

Now,

∴ A is non-singular. 

Therefore, A−1 exists. 

Hence, the given system of equations is consistent.

Answer 7

62. Solve system of linear equations, using matrix method.

5x + 2y = 4

7x + 3y = 5

Answer:

The given system of equations is: 

5x + 2y = 4

7x + 3y = 5

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

Now, 

|A| = 15 - 14 = 1 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

Hence, x = 2 and y = -3.

63. Solve system of linear equations, using matrix method.

2x - y = -2

3x + 4y = 3

Answer:

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

Now,

|A| = 8 + 3 = 11 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

Hence, x = \(\frac{-5}{11}\) and y = \(\frac{12}{11}\).

64. Solve system of linear equations, using matrix method.

4x - 3y = 3

3x - 5y = 7

Answer:

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

Now,

|A| = -20 + 9 = - 11 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Now,

Hence, x = \(\frac{-6}{11}\) and y = \(\frac{-19}{11}\).

65. Solve system of linear equations, using matrix method. 

2x + y + z = 1

x - 2y - z = 3/2

3y - 5z = 9

Answer:

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

Now,

|A| = 2(10 + 3) - 1(-5 - 3) + 0 = 2(13) - 1(-8) = 26 + 8 = 34 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Hence, x = 1, y = \(\frac12\) and z = \(-\frac32\).

66. Solve system of linear equations, using matrix method. 

x − y + z = 4 

2x + y − 3z = 0 

x + y + z = 2

Answer:

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

Now,

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Hence, x = 2, y = -1 and z = 1.

67. Solve system of linear equations, using matrix method. 

2x + 3y + 3z = 5 

x − 2y + z = −4 

3x − y − 2z = 3

Answer:

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

Now,

|A| = 2(4 + 1) - 3(-2 -3) + 3(-1 + 6) = 2(5) - 3(-5) = 10 + 15 + 15 = 40 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

Hence, x = 1, y = 2 and z = -1.

68. Solve system of linear equations, using matrix method. 

x − y + 2z = 7 

3x + 4y − 5z = −5 

2x − y + 3z = 12

Answer:

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

Now,

|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8) = 7 + 19 - 22 = 4 \(\ne\) 0

Thus, A is non-singular. Therefore, its inverse exists.

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

69. If 

, 

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

2x - 3y + 5z = 11

3x + 2y - 4z = -5

x + y - 2z = -3

Answer:

\(\therefore\) |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 \(\ne\) 0

Now, 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^-1B.

X = A^-1B

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

Answer 8

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

Answer:

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs 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

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

Now, 

X = A⁻¹ B

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