# NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

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

⇒ x2 - 36 = 36 - 36
⇒ x2 - 36 = 0
⇒ x2 = 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) + (-x2 – xy + x2 )]
= -(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) k2 | A |
(C) k3 | 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 = [aij] of order n , we can associate a number called the determinant of square matrix A, where aij = (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 (x1 , y1), (x2 , y­2) and (x3 , y3 ) 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 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

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

Solution

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

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

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

Solution

A23 = cofactor of a23 = (-1)2+3 M23 = -7
We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
Δ = a21 A21 + a22 A22 + a23 A23 = 2(7) + 0(7) + 1(-7) = 14 - 7 = 7

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

Solution

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.
∴ Δ = a21 A21 + a22 A22 + a23 A23
= yz(z - y) + zx(x - z) + xy(y - x)
= yz2 - y2 z  +x2 z - xz2 + xy2 - x2 y
= (x2 z - y2 z) + (yz2 - xz2 ) + (xy2 - x2 y)
= z(x2  - y2 ) + z2 (y -x) + xy(y -x)
= z(x - y)(x + y) + z2 (y -x) + xy(y - x)
= (x - y)[zx - zy - z2 - 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 Aij  is Cofactors of aij , then value of Δ is given by
(A) a11 A31+ a12 A32 + a13 A33
(B) a11 A11+ a12 A21 + a13 A31
(C) a21 A11+ a22 A12 + a23 A13
(D) a11 A11+ a21 A21 + a31 A31

Solution
We know that :
∆ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∆ = a21 A21 + a22 A22 + a23 A23
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,
A11 = 10 - 0 = 10 , A12 = -(0 - 0) = 0, A13 = 0 - 0 = 0,
A21 = -(10 - 0) = -10, A22 = 5 - 0 = 5, A23 = -(0 - 0) = 0
A31 = 8 - 6= 2 , A32 = -(4 - 0) = -4, A33 = 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,
A11 = -1 - 0 = -1 , A12 = -(4 - 0) = -4, A13 = 8 - 7 = 1
A21 = -(1 - 6) = 5, A22 = 2 + 21 = 23, A23 = -(4 + 7) = -11
A31 = 0 - 3 = 3 , A32 = -(0 - 12) = 12, A33 = -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, A2 - 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 A2  + aA + bI = 0.
Solution

15. For the matrix A =

show that A3 - 6A2 + 5A + 11I = 0 . Hence, find A-1 .
Solution

A3 - 6A2 + 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)
⇒ A2 - 6A + 5I = -11A-1
⇒ A + aI = -bA-1
⇒ A-1 = -(1/11)[A2 - 6A + 5I]  ...(1)
Now,
A2 - 6A + 5I

16. If A =

verify that A3 - 6A2 + 9A - 4I = 0 and hence find A-1.
Solution

∴ A3 - 6A2 + 9A + 4I = 0
Now,
A3 - 6A2 + 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
⇒ A2 - 6A + 9I = 4A-1
⇒ A + aI = -bA-1
⇒ A-1 = (1/4)[A2 - 6A + 9I]  ...(1)
A2 - 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.
+ 2= 2
2x + 3= 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− y = 5
x + = 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.
3x − 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.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Solution

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| = 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.
5x + 2y = 3

3x + 2y = 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,
A11 = 13, A12 = 5, A13 = 3
A21 = 8, A22 = -10 , A23 = -6
A31 = 1, A32 = 3, A33 = -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,
A11 = 4, A12 = -5, A13 = 1
A21 = 2, A22 = 0 , A23 = -2
A31 = 2, A32 = 5, A33 = 3

13. Solve system of linear equations, using matrix method.
2x + 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,
A11 = 5, A12 = 5, A13 = 5
A21 = 3, A22 = -13 , A23 = 11
A31 = 9, A32 = 1, A33 = -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,
A11 = 7, A12 = -19, A13 = -11
A21 = 1, A22 = -1 , A23 = -1
A31 = -3, A32 = 11, A33 = 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(–x2 – 1) – sinθ(– x sinθ – cos θ) + cos θ(– sinθ + x cosθ)
= –x3 – x + xsin2 θ + sinθ cosθ – sinθ cosθ + x cos2θ
= –x3 – x + x(sin2θ + cos2θ)
= –x3 – x + x
= –x3 (Independent of θ)
Hence, Δ is independent of θ.

2. Without expanding the determinant, prove that

Solution

3. Evaluate

Solution  Δ =

Expanding along C3 , we have :
∆ = − sin α(−sin α sin2β – cos2 β sinα) + cos α(cos α cos2 β + cos α sin2 β)
= sin2 α(sin2 β + cos2 β) + cos2 α(cos2 β + sin2 β)
= sin2 α(1) + cos2 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 R1 , we have :
Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)]
= 2(a + b + c)[-b2 - c2 + 2bc - bc + ba + ac - a2 ]
= 2(a + b + c)[ab + bc +  ca - a2  - b2  - c2 ]

It is given that Δ = 0.
(a + b + c)[ab + bc + ca - a2 - b2 - c2] = 0
⇒ Either a + b + c = 0, or ab + bc + ca - a2 - b2 - c2 = 0 .
Now,
ab + bc + ca - a2 - b2 - c2 = 0
⇒ -2ab - 2bc - 2ca + 2a2 + 2b2 + 2c2  = 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 R1 , we have :
(3x + a)[1 × a2 ]= 0
⇒ a2 (3x + a) = 0
But a ≠ 0.
Therefore, we have :
3x + a = 0
⇒ x = -a/3

6. Prove that

= 4a2 b2 c2
Solution

Expanding along R3 , we have :
Δ = 2ab2 c[a(c - a) + a(a + c)]
= 2ab2 c[ac - a2 + a2 + ac]
= 2ab2 c(2ac)
= 4a2 b2 c2
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 R3 , 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 C1 , we have :
Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]
= (a + b +c)[4abc + 2ab + 2ac + a2 - a2 + 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,
A11 = 75, A12 = 110, A13 = 72
A21 = 150, A22 = -100, A23 = 0
A31 = 75, A32 = 30, A33 = -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 (R1 ) 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 + sin2 θ) - sinθ( - sinθ + sinθ) + 1 (sin2 θ + 1)
= 1 + sin2 θ + sin2 θ + 1
= 2 + 2sin2 θ
= 2(1 + sin2 θ)
Now, 0 ≤ θ ≤ 2π
⇒ 0 ≤ sinθ  ≤ 1
⇒ 0 ≤ sin2 θ  ≤ 1
⇒ 1 ≤ 1 + sin2 θ  ≤ 2
⇒ 2 ≤ 2(1 + sin2 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

## More Study Material

Tags

### NCERT Solutions Class 12 Mathematics Chapter 4 Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 12 Mathematics textbook online or you can easily download them in pdf.

### Chapter 4 Determinants Class 12 Mathematics NCERT Solutions

The Class 12 Mathematics NCERT Solutions Chapter 4 Determinants are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 4 Determinants of Mathematics Class 12 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 4 Determinants Class 12 chapter of Mathematics so that it can be easier for students to understand all answers.

#### NCERT Solutions Chapter 4 Determinants Class 12 Mathematics

Class 12 Mathematics NCERT Solutions Chapter 4 Determinants is a really good source using which the students can get more marks in exams. The same questions will be coming in your Class 12 Mathematics exam. Learn the Chapter 4 Determinants questions and answers daily to get a higher score. Chapter 4 Determinants of your Mathematics textbook has a lot of questions at the end of chapter to test the students understanding of the concepts taught in the chapter. Students have to solve the questions and refer to the step-by-step solutions provided by Mathematics teachers on studiestoday to get better problem-solving skills.

#### Chapter 4 Determinants Class 12 NCERT Solution Mathematics

These solutions of Chapter 4 Determinants NCERT Questions given in your textbook for Class 12 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 12.

#### Class 12 NCERT Solution Mathematics Chapter 4 Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 12 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 12 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 12 Mathematics to clarify all doubts

Where can I download latest NCERT Solutions for Class 12 Mathematics Chapter 4 Determinants

You can download the NCERT Solutions for Class 12 Mathematics Chapter 4 Determinants for latest session from StudiesToday.com

Can I download the NCERT Solutions of Class 12 Mathematics Chapter 4 Determinants in Pdf

Yes, you can click on the link above and download NCERT Solutions in PDFs for Class 12 for Mathematics Chapter 4 Determinants

Are the Class 12 Mathematics Chapter 4 Determinants NCERT Solutions available for the latest session

Yes, the NCERT Solutions issued for Class 12 Mathematics Chapter 4 Determinants have been made available here for latest academic session

How can I download the Chapter 4 Determinants Class 12 Mathematics NCERT Solutions

You can easily access the links above and download the Chapter 4 Determinants Class 12 NCERT Solutions Mathematics for each chapter

Is there any charge for the NCERT Solutions for Class 12 Mathematics Chapter 4 Determinants

There is no charge for the NCERT Solutions for Class 12 Mathematics Chapter 4 Determinants you can download everything free

How can I improve my scores by reading NCERT Solutions in Class 12 Mathematics Chapter 4 Determinants

Regular revision of NCERT Solutions given on studiestoday for Class 12 subject Mathematics Chapter 4 Determinants can help you to score better marks in exams

Are there any websites that offer free NCERT solutions for Chapter 4 Determinants Class 12 Mathematics

Yes, studiestoday.com provides all latest NCERT Chapter 4 Determinants Class 12 Mathematics solutions based on the latest books for the current academic session

Can NCERT solutions for Class 12 Mathematics Chapter 4 Determinants be accessed on mobile devices

Yes, studiestoday provides NCERT solutions for Chapter 4 Determinants Class 12 Mathematics in mobile-friendly format and can be accessed on smartphones and tablets.

Are NCERT solutions for Class 12 Chapter 4 Determinants Mathematics available in multiple languages

Yes, NCERT solutions for Class 12 Chapter 4 Determinants Mathematics are available in multiple languages, including English, Hindi