NCERT Solutions Class 12 Mathematics Chapter 4 Determinants have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 12 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 12 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 12 Mathematics are an important part of exams for Class 12 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 12 Mathematics and also download more latest study material for all subjects. Chapter 4 Determinants is an important topic in Class 12, please refer to answers provided below to help you score better in exams

## Chapter 4 Determinants Class 12 Mathematics NCERT Solutions

Class 12 Mathematics students should refer to the following NCERT questions with answers for Chapter 4 Determinants in Class 12. These NCERT Solutions with answers for Class 12 Mathematics will come in exams and help you to score good marks

### Chapter 4 Determinants NCERT Solutions Class 12 Mathematics

**Exercise 4.1**

**1. Evaluate the determinants in Exercises 1 and 2.**

**Solution**

**2. Evaluate the determinants in Exercises 1 and 2. **

**Solution**

**3. If A**

**Solution**

**4. If A**

**Solution**

**5. Evaluate the determinants **

**Solution**

**6. If A**

**Solution**

**7. Find values of x, if **

**Solution**

**8. If**

**(A) 6
(B) ± 6
(C) -6
(D) 0
Solution**

⇒ x^{2} - 36 = 36 - 36

⇒ x^{2} - 36 = 0

⇒ x^{2} = 36

⇒ x = ± 6

Hence, the correct answer is B.

**Exercise 4.2**

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

**Solution**

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

**Solution**

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

**Solution**

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

**Solution**

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

**Solution**

**6. By using properties of determinants, show that: **

**Solution**

We have,

**7. By using properties of determinants, show that : **

**Solution**

**8. By using properties of determinants, show that : **

**Solution**

= (a - b)(b - c)(c - a)(a + b + c)

Hence, the given result is proved.

**9. By using properties of determinants, show that : **

**Solution**

= (x - y)(z - x)(z- y)[(-xz - yz) + (-x^{2} – xy + x^{2} )]

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

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

Hence, the given result is proved.

**10. By using properties of determinants, show that : **

**Solution**

**11. By using properties of determinants, show that : **

**Solution**

Hance, the guven result is proved

**12. By using properties of determinants show that : **

**Solution**

Hance, the guven result is proved

**13. By using properties of determinants show that : **

**Solution**

**14. By using properties of determinants show that : **

**Solution**

**15. Let A be a square matrix of order 3 × 3, then | kA| is equal to
(A) k|A|
(B) k**

^{2}| A | (C) k

^{3}| A | (D) 3k | A |

**Solution**

**6. 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 the above.
Solution**

We know that to every square matrix, A = [a

_{ij}] of order n , we can associate a number called the determinant of square matrix A, where a

_{ij}= (i, j)

^{th}element of A. Thus, the determinant is a number associated to a square matrix. Hence, the correct option is C.

**Exercise 4.3**

**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)
Solution**

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) 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]

= -30/2 = -15

Hence, area of the triangle is 15 square units.

**2. Show that the points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
Solution**

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 A, B, and C is zero.

Hence, the points A, B and C 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)
Solution**

We know that the area of a triangle whose vertices are (x

_{1}, y

_{1}), (x

_{2}, y

_{2}) and (x

_{3}, y

_{3}) is the absolute value of the determinant (Δ), where

= (1/2) [-2(4 - k)

= k - 4

∴ k - 4 = ± 4

When k - 4 = - 4 , k = 0

When k - 4 = 4, k = 8

Hence, k = 0, 8

**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.
Solution**

(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.

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

⇒ 6 - y - 6 + 2x + 3y - 6x = 0

⇒ 2y - 4x = 0

⇒ y = 2x

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.

⇒ 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

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

**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

Solution

The area of the triangle with vertices (2, -6), (5, 4), and (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.

Therefore, we have :

⇒ 25 - 5k = ± 35

⇒ 5(5 - k) = ± 35

⇒ 5 - k = ± 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.

**Exercise 4.4**

**1. Write Minors and Cofactors of the elements of following determinants : **

**Solution**

(i) The given determinant is

Minor of a_{ij} is M_{ij} .

M_{11} = minor of element a_{11} = 3

M_{12} = minor of element a_{12} = 0

M_{21} = minor of element a_{21} = -4

M_{22} = minor of element a_{22} = 2

Cofactor of a_{ij} is A_{ij} = (-1)^{i+j} M_{ij} .

∴ A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} (3) = 3

A_{12} = (-1)^{1+2} M_{12} = (-1)^{3} (0) = 0

A_{21} = (-1)^{2+1} M_{21} = (-1)^{3} (-4) = 4

A_{22} = (-1)^{2+2} M_{22} = (-1)^{4} (2) = 2

(ii) The given determinant is

Minor of element a_{ij} is M_{ij} .

∴ M_{11} = minor of element a_{11} = d

M_{12} = minor of element a_{12} = b

M_{21} = minor of element a_{21} = c

M_{22} = minor of element a_{22} = a

Cofactor of a_{ij} is A_{ij} = (-1)^{i+j} M_{ij} .

∴ A_{11} = (-1)^{1+1} M_{11} = (-1)^{2} (d) = d

A_{12} = (-1)^{1+2} M_{12} = (-1)^{3} (b) = -b

A_{21} = (-1)^{2+1} M_{21} = (-1)^{3} (c) = -c

A_{22} = (-1)^{2+2} M_{22} = (-1)^{4} (a) = a

**2. Write Minors and Cofactors of the elements of following determinants: **

**Solution**

A_{11} = cofactor of a_{11} = (-1)^{1+1} M_{11} = 1

A_{12} = cofactor of a_{12} = (-1)^{1+2} M_{12} = 0

A_{13} = cofactor of a_{13} = (-1)^{1+3} M_{13} = 0

A_{21} = cofactor of a_{21} = (-1)^{2+1} M_{21} = 0

A_{22} = cofactor of a_{22} = (-1)^{2+2} M_{22} = 1

A_{23} = cofactor of a_{23} = (-1)^{2+3} M_{23} = 0

A_{31} = cofactor of a_{31} = (-1)^{3+1} M_{31} = 0

A_{32} = cofactor of a_{32} = (-1)^{3+2} M_{32} = 0

A_{33} = cofactor of a_{33} = (-1)^{3+3} M_{33} = 1

A_{11} = cofactor of a_{11} = (-1)^{1+1} M_{11} = 11

A_{12} = cofactor of a_{12} = (-1)^{1+2} M_{12} = -6

A_{13} = cofactor of a_{13} = (-1)^{1+3} M_{13} = 3

A_{21} = cofactor of a_{21} = (-1)^{2+1} M_{21} = 4

A_{22} = cofactor of a_{22} = (-1)^{2+2} M_{22} = 2

A_{23} = cofactor of a_{23} = (-1)^{2+3} M_{23} = -1

A_{31} = cofactor of a_{31} = (-1)^{3+1} M_{31} = -20

A_{32} = cofactor of a_{32} = (-1)^{3+2} M_{32} = 13

A_{33} = cofactor of a_{33} = (-1)^{3+3} M_{33} = 5

**3. Using Cofactors of elements of second row, evaluate Δ =**

**Solution**

A_{23} = cofactor of a_{23} = (-1)^{2+3} M_{23} = -7

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

Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = 2(7) + 0(7) + 1(-7) = 14 - 7 = 7

**4. sing Cofactors of elements of third column, evaluate Δ =**

**Solution**

A_{13} = cofactor of a_{13} = (-1)^{1+3} M_{13} = (z- y)

A_{23} = cofactor of a_{23} = (-1)^{2+3} M_{23} = -(z - x) = (x - z)

A_{33} = cofactor of a_{33} = (-1)^{3+3} M_{33} = (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.

∴ Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}

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

= yz^{2} - y^{2} z +x^{2} z - xz^{2} + xy^{2} - x^{2} y

= (x^{2} z - y^{2} z) + (yz^{2} - xz^{2} ) + (xy^{2} - x^{2} y)

= z(x^{2} - y^{2} ) + z^{2} (y -x) + xy(y -x)

= z(x - y)(x + y) + z^{2} (y -x) + xy(y - x)

= (x - y)[zx - zy - z^{2} - xy]

= (x - y) [z(x -z) + y(z - x)]

= (x - y) (z - x)[-z + y]

= (x - y)(y - z)(z - x)

Hence, Δ = (x - y)(y - z)(z - x) .

**5. If ∆ = **

**and A _{ij} is Cofactors of a_{ij} , then value of Δ is given by**

(A) a_{11} A_{31}+ a_{12} A_{32} + a_{13} A_{33}

(B) a_{11} A_{11}+ a_{12} A_{21} + a_{13} A_{31}

(C) a_{21} A_{11}+ a_{22} A_{12} + a_{23} A_{13}

(D) a_{11} A_{11}+ a_{21} A_{21} + a_{31} A_{31}

**Solution**

We know that :

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

∆ = a

_{21}A

_{21}+ a

_{22}A

_{22}+ a

_{23}A

_{23}

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

The correct answer is D.

**Exercise 4.5**

**1. Find the adjoint of the matrix**

**Solution**

**2. Find adjoint of each of the matrices.**

**Solution**

**3. Verify A (adj A) = (adj A) A = |A|**

**Solution**

**4. Verify A (adj A) = (adj A) A = |A|**

**Solution**

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

**Solution**

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

**Solution**

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

**Solution**

We have,

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

Now,

A_{11} = 10 - 0 = 10 , A_{12} = -(0 - 0) = 0, A_{13} = 0 - 0 = 0,

A_{21} = -(10 - 0) = -10, A_{22} = 5 - 0 = 5, A_{23} = -(0 - 0) = 0

A_{31} = 8 - 6= 2 , A_{32} = -(4 - 0) = -4, A_{33} = 2 - 0 = 2

**8. Find the inverse of each of the matrix (if it exists). **

**Solution**

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

**Solution**

We have,

|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)

= 2(-1) - 1(4) + 3(1)

= -2 - 4 + 3

= - 3

Now,

A_{11} = -1 - 0 = -1 , A_{12} = -(4 - 0) = -4, A_{13} = 8 - 7 = 1

A_{21} = -(1 - 6) = 5, A_{22} = 2 + 21 = 23, A_{23} = -(4 + 7) = -11

A_{31} = 0 - 3 = 3 , A_{32} = -(0 - 12) = 12, A_{33} = -2 - 4 = -6

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

**Solution**

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

**Solution**

**12. Let A =**

**Solution**

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

(AB)^{-1} = B^{-1} A^{-1}

Hence, the given result is proved.

**13. If A =**

**Solution**

Hence, A^{2} - 5A + 7I = 0.

∴ A . A - 5A = -7I

⇒ A.A(A^{-1} ) - 5AA^{-1} = -7IA^{-1} [Post-multiplying by A^{-1} as |A| ≠ 0]

⇒ A(AA^{-1} ) - 5I = -7A^{-1}

⇒ AI -5I = -7A^{-1}

⇒ A^{-1} = - (1/7)[A - 5I]

⇒ A^{-1} = (1/7)[5I - A]

**14. For the matrix A = **

**find the numbers a and b such that A ^{2} + aA + bI = 0.
Solution**

**15. For the matrix A =**

**show that A ^{3} - 6A^{2} + 5A + 11I = 0 . Hence, find A^{-1} . **

Solution

A^{3} - 6A^{2} + 5A + 11I = 0

⇒ (AAA)A^{-1} - 6(AA)A^{-3} + 5AA^{-1} + 11IA^{-1} = 0 [Post-multiplying by A^{-1} as |A| ≠ 0]

⇒ AA(AA^{-1} ) - 6A(AA^{-1} ) + 5(AA^{-1} ) = 11(IA^{-1})

⇒ A^{2} - 6A + 5I = -11A^{-1}

⇒ A + aI = -bA^{-1}

⇒ A^{-1} = -(1/11)[A^{2} - 6A + 5I] **...(1)**

Now,

A^{2} - 6A + 5I

**16. If A = **

**verify that A ^{3} - 6A^{2} + 9A - 4I = 0 and hence find A^{-1}.**

Solution

∴ A^{3} - 6A^{2} + 9A + 4I = 0

Now,

A^{3} - 6A^{2} + 9A + 4I = 0

⇒ (AAA)A^{-1} - 6(AA)A^{-3} + 9AA^{-1} + 4IA^{-1} = 0 [Post-multiplying by A^{-1} as |A| ≠ 0]

⇒ AA(AA^{-1} ) - 6A(AA^{-1} ) + 9(AA^{-1} ) = 4(IA^{-1} )

⇒ AAI - 6AI + 9I = 4A^{-1}

⇒ A^{2} - 6A + 9I = 4A^{-1}

⇒ A + aI = -bA^{-1}

⇒ A^{-1} = (1/4)[A^{2} - 6A + 9I] **...(1)**

A^{2} - 6A + 9I

**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| Solution

We know that,

**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

Solution

Since A is an invertible matrix, A

^{-1}exists and A

^{-1}= (1/|A|) adj A.

**Exercise 4.6**

**1. Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Solution**

The given system of equation 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 ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**2. Examine the consistency of the system of equations.
2 x − y = 5
x + y = 4
Solution**

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 ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**3. Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solution**

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

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

**4. Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution**

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 ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**5. Examine the consistency of the system of equations.
3 x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Solution**

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

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

**6. Examine the consistency of the system of equations.
5 x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Solution**

The given system of equations is :

5

*x*−

*y*+ 4

*z*= 5

2

*x*+ 3

*y*+ 5

*z*= 2

5

*x*− 2

*y*+ 6

*z*= −1

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

Now,

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

= 5(28) + 1(-13) + 4(-19)

= 140 - 13 - 76

= 51 ≠ 0

A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**7. Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Solution**

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

**8. Solve system of linear equations, using matrix method.
2x – y = –2
3x + 4y = 3
Solution**

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

**9. Solve system of linear equations, using matrix method.
4x – 3y = 3
3x – 5y = 7
Solution**

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

**10.Solve system of linear equations, using matrix method.
5 x + 2y = 3**

**3**

*x*+ 2*y*= 5**Solution**

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

**11. Solve system of linear equations, using matrix method.
2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9
Solution**

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 ≠ 0

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

Now,

A_{11} = 13, A_{12} = 5, A_{13} = 3

A_{21} = 8, A_{22} = -10 , A_{23} = -6

A_{31} = 1, A_{32} = 3, A_{33} = -5

**12. Solve system of linear equations, using matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Solution**

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 ≠ 0

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

Now,

A_{11} = 4, A_{12} = -5, A_{13} = 1

A_{21} = 2, A_{22} = 0 , A_{23} = -2

A_{31} = 2, A_{32} = 5, A_{33} = 3

**13. Solve system of linear equations, using matrix method.
2 x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Solution**

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) + 3(5)

= 10 + 15 + 15

= 40 ≠ 0

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

Now,

A_{11} = 5, A_{12} = 5, A_{13} = 5

A_{21} = 3, A_{22} = -13 , A_{23} = 11

A_{31} = 9, A_{32} = 1, A_{33} = -7

**14. Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Solution**

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 ≠ 0

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

Now,

A_{11} = 7, A_{12} = -19, A_{13} = -11

A_{21} = 1, A_{22} = -1 , A_{23} = -1

A_{31} = -3, A_{32} = 11, A_{33} = 7

**15. If A = **

**find A^{−1}. Using A^{−1} solve the system of equations**

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

Solution

**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.
Solution**

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

x = 5, y = 8 and z = 8

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

**Miscellaneous Solutions**

**1. Prove that the determinant **

**is independent of θ is independent of** **θ.
Solution**

= x(–x^{2} – 1) – sinθ(– x sinθ – cos θ) + cos θ(– sinθ + x cosθ)

= –x^{3} – x + xsin^{2} θ + sinθ cosθ – sinθ cosθ + x cos^{2}θ

= –x^{3} – x + x(sin^{2}θ + cos^{2}θ)

= –x^{3} – x + x

= –x^{3} (Independent of θ)

Hence, Δ is independent of θ.

**2. Without expanding the determinant, prove that**

**Solution**

**3. Evaluate **

**Solution** Δ =

Expanding along C_{3} , we have :

∆ = − sin α(−sin α sin^{2}β – cos^{2} β sinα) + cos α(cos α cos^{2} β + cos α sin^{2} β)

= sin^{2} α(sin^{2} β + cos^{2} β) + cos^{2} α(cos^{2} β + sin^{2} β)

= sin^{2} α(1) + cos^{2} α(1)

= 1

**4. If a, b and c are real numbers, and triangle = **

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

**Solution**

Expanding along R_{1} , we have :

Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)]

= 2(a + b + c)[-b^{2} - c^{2} + 2bc - bc + ba + ac - a^{2} ]

= 2(a + b + c)[ab + bc + ca - a^{2} - b^{2} - c^{2} ]

It is given that Δ = 0.

(a + b + c)[ab + bc + ca - a^{2} - b^{2} - c^{2}] = 0

⇒ Either a + b + c = 0, or ab + bc + ca - a^{2} - b^{2} - c^{2} = 0 .

Now,

ab + bc + ca - a^{2} - b^{2} - c^{2} = 0

⇒ -2ab - 2bc - 2ca + 2a^{2} + 2b^{2} + 2c^{2} = 0

⇒ (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0

⇒ (a - b)^{2} = (b - c)^{2} = (c - a)^{2} = 0 [(a-b)^{2}, (b-c)^{2}, (c-a)^{2} are non-negative]

⇒ (a - b) = (b - c) = (c - a) = 0

⇒ a = b = c

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

**5. Solve the equations**

**= 0, a ≠ 0
Solution**

Expanding along R_{1} , we have :

(3x + a)[1 × a^{2} ]= 0

⇒ a^{2} (3x + a) = 0

But a ≠ 0.

Therefore, we have :

3x + a = 0

⇒ x = -a/3

**6. Prove that **

**= 4a ^{2} b^{2} c^{2} **

Solution

Expanding along R_{3} , we have :

Δ = 2ab^{2} c[a(c - a) + a(a + c)]

= 2ab^{2} c[ac - a^{2} + a^{2} + ac]

= 2ab^{2} c(2ac)

= 4a^{2} b^{2} c^{2}

Hence, the given result is proved.

**7. If A ^(-1) = **

**Solution**

We know that (AB)^{-1} = B^{-1} A^{-1} .

**8. Let A = **

**verify that
(i) [adj A] ^{-1} = adj(A^{-1})
(ii) (A^{-1})^{-1} = A
Solution**

(i) |adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20)

= 14(-13) - 11(-26) - 5(-13)

= -182 + 286 + 65 = 169

we have ,

**9. Evaluate**

**Solution**

**10. Evaluate **

**Solution**

**11. Using properties of determinants, prove that : **

**Solution**

Expanding along R_{3} , we have :

Δ = (β - α) (γ - α)[- (γ - β)( - α - β - γ)]

= (β - α) (γ - α) (γ - β)(α + β + γ)

= (α - β) (β - γ) (γ - α)(α + β + γ)

Hence, the given result is proved.

**12. Using properties of determinants, prove that : **

**(1 + pxyz)(x - y)(y - z)(z -x), where p is any scalar.
Solution**

**13. Using properties of determinants, prove that :**

**Solution**

Expanding along C_{1} , we have :

Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]

= (a + b +c)[4abc + 2ab + 2ac + a^{2} - a^{2} + ac + ba - bc]

= (a + b + c)(3ab + 3bc + 3ac)

= 3(a + b + c)(ab + bc + ca)

Hence, the given result is proved.

**14. Using properties of determinants, prove that : **

**Solution**

**15 . Solve the system of the following equations
2/x + 3/y + 10/z = 4
4/x - 6/y + 5/z = 1
6/x + 9/y - 20/x = 2
Solution**

Let 1/x = p, 1/y = q, 1/z = r.

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

Now,

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

= 150 + 330 + 720

=1200

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

Now,

A_{11} = 75, A_{12} = 110, A_{13} = 72

A_{21} = 150, A_{22} = -100, A_{23} = 0

A_{31} = 75, A_{32} = 30, A_{33} = -24

∴ A^{-1} = 1/|A| [adj A]

**16. Choose the correct answer.
If a, b, c, are in A.P., then the determinant **

**A. 0
B. 1
C. x
D. 2x
Solution**

Here, all the elements of the first row (R_{1} ) are zero.

Hence, we have Δ = 0.

The correct answer is A.

**18. If x, y, z are nonzero real numbers, then the inverse of matrix A =**

**Solution**

**19. Choose the correct answer . **

**A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A)∈ [2, 4]
Solution**

∴ |A| = 1(1 + sin^{2} θ) - sinθ( - sinθ + sinθ) + 1 (sin^{2} θ + 1)

= 1 + sin^{2} θ + sin^{2} θ + 1

= 2 + 2sin^{2} θ

= 2(1 + sin^{2} θ)

Now, 0 ≤ θ ≤ 2π

⇒ 0 ≤ sinθ ≤ 1

⇒ 0 ≤ sin^{2} θ ≤ 1

⇒ 1 ≤ 1 + sin^{2} θ ≤ 2

⇒ 2 ≤ 2(1 + sin^{2} θ) ≤ 2

∴ Det (A) ∊ [2, 4]

The correct answer is D.

NCERT Solutions Class 12 Mathematics Chapter 1 Relations and Functions |

NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions |

NCERT Solutions Class 12 Mathematics Chapter 3 Matrices |

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants |

NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability |

NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives |

NCERT Solutions Class 12 Mathematics Chapter 7 Integrals |

NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals |

NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations |

NCERT Solutions Class 12 Mathematics Chapter 10 Vector Algebra |

NCERT Solutions Class 12 Mathematics Chapter 11 Three Dimensional Geometry |

NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming |

NCERT Solutions Class 12 Mathematics Chapter 13 Probability |

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