NCERT Solutions for Class 12 Maths Chapter 4 “Determinants” is a crucial chapter for students who wish to pursue higher mathematics. In this chapter, students will learn about determinants, which are a mathematical tool used to solve linear equations, find the inverse of matrices, and solve a variety of mathematical problems.
The chapter begins by introducing the concept of determinants and its properties. Students will learn how to calculate determinants of matrices of different sizes and the properties of determinants, such as the multiplicative property and the triangular property.
NCERT Solutions for Class 12 Maths Chapter 4 also provide detailed explanations and a variety of solved examples on how to evaluate determinants using different methods such as expansion along rows or columns, minors and cofactors, and properties of determinants. Additionally, the solutions offer step-by-step guidance on how to approach different types of problems and help students to develop problem-solving skills.

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Answers of Mathematics NCERT solutions for class 12 Chapter 4 Determinants

Chapter 4

Determinants

Exercise 4.1

Question 1:

Evaluate the determinants in Exercises 1 and 2.

Answer:

 = 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2:

Evaluate the determinants

(i)  (ii) 

Answer:

(i)   

= (cos θ)(cos θ) − (−sin θ)(sin θ) = cos² θ+ sin² θ = 1

(ii)  

= (x² − x + 1)(x + 1) − (x − 1)(x + 1)

= x³ − x² + x + x² − x + 1 − (x² − 1)

= x³ + 1 − x² + 1

= x³ − x² + 2

Question 3:

If then show that

Answer:

The given matrix is

Question 4:

If then show that

Answer:

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.

From equations (i) and (ii), we have:

Hence, the given result is proved.

Question 5:

Evaluate the determinants

Answer:

(i) Let

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

(ii) Let

By expanding along the first row, we have:

(iii) Let

By expanding along the first row, we have:

(iv) Let

By expanding along the first column, we have:

Question 6:

If , find

Answer:

Let

By expanding along the first row, we have:

Question 7:

Find values of x, if

(i)

(ii)

Answer:

(i) 

(ii) 

Question 8:

If then x is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

Answer: B

Hence, the correct answer is B.

Exercise 4.2

*Question 1:

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

Answer:

Question 2:

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

Answer:

Here, the two rows R₁ and R₃ are identical.

Δ = 0.

Question 3:

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

Answer:

*Question 4:

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

Answer:

By applying C₃ → C₃ + C_2, we have:

Here, two columns C₁ and C₃ are proportional.

Δ = 0.

Question 5:

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

Answer:

Applying R₂ → R₂ − R₃, we have:

Applying R₁ ↔R₃ and R₂ ↔R₃, we have:

Applying R₁ → R₁ − R₃, we have:

Applying R₁ ↔R₂ and R₂ ↔R₃, we have:

From (1), (2), and (3), we have:

Hence, the given result is proved.

Question 6:

By using properties of determinants, show that:

Answer:

We have,

Here, the two rows R₁ and R₃ are identical.

∴Δ = 0.

Question 7:

By using properties of determinants, show that:

Answer:

Applying R₂ → R₂ + R₁ and R₃ → R₃ + R₁, we have:

Question 8:

By using properties of determinants, show that:

(i)  

(ii) 

Answer:

(i) 

Applying R₁ → R₁ − R₃ and R₂ → R₂ − R₃, we have:

Applying R₁ → R₁ + R₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

(ii) Let .

Applying C₁ → C₁ − C₃ and C₂ → C₂ − C₃, we have:

Applying C₁ → C₁ + C₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

*Question 9:

By using properties of determinants, show that:

Answer:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying R₃ → R₃ + R₂, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 10:

By using properties of determinants, show that:

(i) 

(ii)  

Answer:

(i) 

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii) 

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

Question 11:

By using properties of determinants, show that:

(i)  

(ii) 

Answer:

(i) 

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii)  

Applying C₁ → C₁ + C₂ + C₃, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 12:

By using properties of determinants, show that:

Answer:

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along R₁, we have:

Hence, the given result is proved.

*Question 13:

By using properties of determinants, show that:

Answer:

Applying R₁ → R₁ + bR₃ and R₂ → R₂ − aR₃, we have:

Expanding along R₁, we have:

Question 14:

By using properties of determinants, show that:

Answer:

Taking out common factors a, b, and c from R₁, R₂, and R₃ respectively, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying C₁ → aC₁, C₂ → bC_2, and C₃ → cC₃, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 15:

Choose the correct answer.

Let A be a square matrix of order 3 × 3, then  is equal to

A.     B.     C.   D.

Answer: C

A is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Question 16:

Which of 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: C

We know that to every square matrix,  of order n. We can associate a number called the determinant of square matrix A, where  element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.

Exercise 4.3

Question 1:

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 square units.

*Question 2:

Show that points

      are 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.

*Question 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)

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.

Question 4:

(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.

Question 5:

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: D

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:

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.

Exercise 4.4 

Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i)   (ii) 

Answer:

(i) The given determinant is .

Minor of element aij is Mij.

∴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 aij is Aij = (−1)i + j Mij.

∴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 aij is Mij.

∴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 aij is Aij = (−1)i + j Mij.

∴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

Question 2:

(i)   (ii) 

Answer:

(i) The given determinant is .

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁= 

M₁₂ = minor of a₁₂= 

M₁₃ = minor of a₁₃ = 

M₂₁ = minor of a₂₁ = 

M₂₂ = minor of a₂₂ = 

M₂₃ = minor of a₂₃ = 

M₃₁ = minor of a₃₁= 

M₃₂ = minor of a₃₂ = 

M₃₃ = minor of a₃₃ = 

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 1

A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = 0

A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 0

A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 0

A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 1

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = 0

A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = 0

A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 0

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 1

(ii) The given determinant is .

By definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁= 

M₁₂ = minor of a₁₂= 

M₁₃ = minor of a₁₃ = 

M₂₁ = minor of a₂₁ = 

M₂₂ = minor of a₂₂ = 

M₂₃ = minor of a₂₃ = 

M₃₁ = minor of a₃₁= 

M₃₂ = minor of a₃₂ = 

M₃₃ = minor of a₃₃ = 

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 11

A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = −6

A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 3

A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 4

A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 2

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −1

A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = −20

A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 13

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 5

Question 3:

Using Cofactors of elements of second row, evaluate

Answer:

The given determinant is

We have:

M₂₁ = 

∴A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 7

M₂₂ = 

∴A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 7

M₂₃ = 

∴A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −7

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

∴Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Question 4:

Using Cofactors of elements of third column, evaluate

Answer:

The given determinant is.

We have:

M₁₃ = 

M₂₃ = 

M₃₃ = 

∴A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = (z − y)

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = − (z − x) = (x − z)

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = (y − x)

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

Hence, 

Question 5:

If  and Aij is Cofactors of aij, then value of Δ is given by

Answer: D

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.

Exercise 4.5

Question 1:

Find adjoint of each of the matrices.

Answer:

Question 2:

Find adjoint of each of the matrices.

Answer:

Question 3:

Verify A (adj A) = (adj A) A = |A|I .

Answer:

Question 4:

Verify A (adj A) = (adj A) A = |A|I .

Answer:

Question 5:

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

Answer:

Question 6:

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

Answer:

Question 7:

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

Answer:

*Question 8:

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

Answer:

Question 9:

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

Answer:

Question 10:

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

.

Answer:

*Question 11:

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

Answer:

Question 12:

Let  and Verify that 

Answer:

From (1) and (2), we have:

(AB)⁻¹ = B⁻¹A⁻¹

Hence, the given result is proved.

*Question 13:

If show that Hence find

Answer:

Question 14:

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

Answer:

We have:

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

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

Question 15:

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

Answer:

From equation (1), we have:

Question 16:

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

Answer:

From equation (1), we have:

Question 17:

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

A.   B.   C.   D. 

Answer: B

We know that,

Hence, the correct answer is B.

Question 18:

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

A. det (A)  B.    C. 1  D. 0

Answer:

Since A is an invertible matrix,  exists and

Hence, the correct answer is B.

Exercise 4.6

Question 1:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 2:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 3:

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

∴ A is a singular matrix.

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

Question 4:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 5:

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

∴ A is a singular matrix.

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

Question 6:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Question 7:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 8:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 9:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 10:

Solve system of linear equations, using matrix method.

5x + 2y = 3

3x + 2y = 5

Answer:

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

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

Question 11:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 12:

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

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

Question 13:

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

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

Question 14:

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

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

*Question 15:

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

Answer:

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

*Question 16:

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:

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.

Miscellaneous Exercise

*Question 1:

Prove that the determinant is independent of θ.

Answer:

Hence, Δ is independent of Î¸.

Question 2:

Without expanding the determinant, prove that

Answer:

Hence, the given result is proved.

Question 3:

Answer:

Expanding along C₃, we have:

Question 4:

If a, b and c are real numbers, and

Show that either a + b + c = 0 or a = b = c.

Answer:

Expanding along R₁, we have:

Hence, if Δ = 0, then either a + b + c = 0 or a = b = c.

Question 5:

Solve the equations 

Answer:

Question 6:

Prove that 

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 7:

If 

Answer:

We know that

Question 8:

Let  verify that

(i) 

(ii) 

Answer:

(i)

We have,

(ii)

Question 9:

Evaluate 

Answer:

Question 10:

Evaluate 

Answer:

Expanding along C₁, we have:

Question 11:

Using properties of determinants, prove that:

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

*Question 12:

Using properties of determinants, prove that:

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

Question 13:

Using properties of determinants, prove that:

Answer:

Expanding along C₁, we have:

Hence, the given result is proved.

Question 14:

Using properties of determinants, prove that:

Answer:

Expanding along C₁, we have:

Hence, the given result is proved.

Question 15:

Using properties of determinants, prove that:

Answer:

Hence, the given result is proved.

*Question 16:

Solve the system of the following equations

Answer:

Let 

Then the given system of equations is as follows:

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

A

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

Now,

A₁₁ = 75, A₁₂ = 110, A₁₃ = 72

A₂₁ = 150, A₂₂ = −100, A₂₃ = 0

A₃₁ = 75, A₃₂ = 30, A₃₃ = − 24

Question 17:

Choose the correct answer.

If a, b, c, are in A.P., then the determinant

A.   0  B.   1  C.   x  D.   2x

Answer: A

Here, all the elements of the first row (R₁) are zero.

Hence, we have Δ = 0.

The correct answer is A.

Question 18:

Choose the correct answer.

If x, y, z are nonzero real numbers, then the inverse of matrix is

A.    B. 

C.    D. 

Answer: A

The correct answer is A.

*Question 19:

Choose the correct answer.

Let , where 0 ≤ θ≤ 2π, then

A.  Det (A) = 0

B.  Det (A) ∈ (2, ∞)

C.  Det (A) ∈ (2, 4)

D.  Det (A)∈ [2, 4]

Answer:

Answer: D

Now,

0 ≤ θ ≤ 2π

 The correct answer is D.

Conclusion

NCERT Solutions for Class 12 Math Chapter 4 “Determinants” provide students with a comprehensive understanding of the concept of determinants and its properties. The chapter covers a variety of topics such as calculation of determinants, properties of determinants, and different methods to evaluate determinants.
The solutions offer clear explanations and numerous solved examples to help students grasp the concepts more effectively. The step-by-step guidance provided in the solutions also assists students in developing problem-solving skills.
Overall, NCERT Solutions for Class 12 Math Chapter 4 is an essential resource for students who want to excel in higher mathematics and develop their skills in solving complex mathematical problems. These solutions help students to build a strong foundation in the topic of determinants, which is a fundamental tool used in various fields of mathematics and science.

Topics to study in Maths Class 12 Chapter 4

Section no	Topics
4.1	Introduction
4.2	Determinant
4.3	Properties for Determinants
4.4	Area of a Triangle
4.5	Minors and Cofactors
4.6	Adjoint and Inverse of a Matrix
4.7	Applications of Determinants and Matrices

Weightage of Determinants in exam – TERM I

Chapters	Marks
Determinants	6 Marks

Related Links

NCERT Solution for Class XIIth Maths Chapter 3 Matrices	NCERT Solution for Class XIIth Maths Chapter 2 Inverse Trigonometry
NCERT Solution for Class XIIth Maths Chapter 5 Continuity and Differentiability	NCERT Solution for Class XIIth Maths Chapter 1 Relations and Function
NCERT Solution for Class XIIth Maths Chapter 7 Integrals	NCERT Solution for Class XIIth Maths Chapter 6 Applications of Derivatives

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Videos on NCERT Maths Class 12 Chapter 4

Frequently Asked Questions on NCERT Solutions for Class 12 Maths Chapter 4

How many problems are there in NCERT Solutions for Class 12 Maths Chapter 4?

Exercise 4.1 of NCERT Solutions has 8 questions, Exercise 4.2 has 16 questions, Exercise 4.3 and 4.4 has 5 questions, Exercise 4.5 has 18 questions, Exercise 4.6 has 16 questions and miscellaneous exercise has 19 questions. Long answers, short answers and MCQs are present in each exercise covering the topics which are important from the exam point of view. The solutions for the exercise wise problems are designed by a set of experts at BYJU’S having vast conceptual knowledge.

What are the uses of determinants according to NCERT Solutions for Class 12 Maths Chapter 4?

Determinants is one of the main concepts having a number of applications in algebra. This concept is useful to solve a set of linear equations. Using determinants, students will be able to understand the change in area, volume, variables in terms of integrals. It can also be used to calculate the values of square matrices.

What kind of problems can I expect in the board exams from NCERT Solutions for Class 12 Maths Chapter 4?

The problems based on solving the given set of linear equations using determinants can be expected from this chapter. Proofs that can be derived using the theory of determinants and prove that sums are also important from the exam perspective. In certain problems, you should determine the values of unknown variables using determinants and their resultant values.

How many exercise determinants are there in class 12?

There are 6 Exercises in class 12 math determinants.

What do you mean by determinants in mathematics?

The determinant is a scalar variable in mathematics that is a function of the elements of a square matrix. It allows you to characterize some of the matrix’s attributes as well as the linear map that the matrix represents.

How do you find the minor of a matrix?

The determinant of each smaller matrix, produced by eliminating the appropriate rows and columns of each element in the matrix, is used to compute the minor of a matrix. We can create several minors of those matrices since there are many rows and columns with many elements in huge matrices.

What is the determinant formula?

The determinant is: |A| = a (ei − fh) − b (di − fg) + c (dh − eg). The determinant of A equals ‘a times e x i minus f x h minus b times d x i minus f x g plus c times d x h minus e x g’.

What if the determinant is 0?

If the determinant is zero, the matrix is not invertible and hence does not have a solution since one of the rows in the matrix may be deleted by substituting another row in the matrix.

How do you solve a determinant equation?

A matrix is often used to represent the coefficients in a system of linear equations, and the determinant can be used to solve those equations.