NCERT Solutions for Class 12 Maths Chapter 4: Determinants
NCERT Solutions for Class 12 Maths Chapter 4 explains the topic of determinants, which forms an essential part of the theory of matrices and linear algebra. The solutions are designed to exemplify the properties and theorems pertaining to determinants, their use in calculating the area of a triangle, methods of solving linear equations, and so on. The Class 12 Maths Chapter 4 PDF explains how to solve determinant-related problems with a step-by-step approach so that students understand the concept clearly.
Download PDF For NCERT Solutions for Maths Determinants
The NCERT Solutions for Class 12 Maths Chapter 4: Determinants are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 12 Maths Chapter 4: Determinants
Students can access the NCERT Solutions for Class 12 Maths Chapter 4: Determinants. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 4.1
Evaluate the following determinants
Evaluate the following determinants
(i)
(ii)
(i)
= (cosθ)(cosθ) – (-sinθ) (sinθ)= 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
If A = 
Given: A =
then 2A = 2 x
Hence, proved.
If A = 
Given: A = 
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.
Hence, proved.
Evaluate the determinants:
(i)
(ii) 
(iii)
(iv) 
Evaluate the determinants:
(i) Given:
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.
=
(ii) Given:
By expanding along the first row, we have:
=
(iii) Given:
Expanding along first row,
=
= 0 + 6 – 6 = 0
(iv) Given:
Expanding along first row,
=
= -10 + 15 = 5
If A = 
Given: A =
Expanding along first row,
=
Find the value of x if:
(i)
(ii)
(i) Given:
⇒ 2 x 1 – 5 x 4 = 2x * x – 6 x 4
⇒ 2 – 20 = 2x2 – 24
⇒ 2x2 = 6
⇒ x2 = 3
⇒ x = ± √3
(ii)
⇒ 2 x 5 – 4 x 3 = x * 5 – 2x – 3
⇒10 – 12 = 5x – 6x
⇒ – 2 = -x
⇒ x = 2
If 
(A) 6
(B) ± 6
(C) – 6
(D) 0
Given:
⇒x * x – 18 x 2 = 6 x 6 – 18 x 2
⇒x2 – 36 = 36 – 36
⇒x2 – 36 = 0
⇒x = ± 6
Therefore, option (B) is correct.
Exercise 4.2
[Here, two columns of the determinants are identical]
On Operating
(i)
(ii)
(i)
(ii)
Let A be a square matrix of order 3 x 3, then k |A| is equal to:
(A) k |A|
(B) k2 |A|
(C) k3 |A|
(D) 3k |A|
Therefore, option (C) is correct.
=
(i)
(ii)
(i)
(ii)
Let A be a square matrix of order 3 x 3, then k |A| is equal to:
(A) k |A|
(B) k2 |A|
(C) k3 |A|
(D) 3k |A|
Exercise 4.3
Find the area of the triangle with vertices at the points 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)
(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)
=
Show that the points A(a,b + c), B(b, c + a), C(c, a+b) are collinear.
Therefore, points A, B and C are collinear.
Find values of k if area of triangle is 4 sq. units and vertices are:
(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)
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
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.
(i) Find the equation of the line joining (1, 2) and (3, 6) using determinants.
(ii) Find the equation of the line joining (3, 1) and (9, 3) using determinants.
(i) Let P(x, y) be any point on the line joining the points (1, 2) and (3, 6).
Then, Area of the triangle that could be formed by these points is 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.
If area of the 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
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 = ± 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.
Therefore, option (D) is correct.
Exercise 4.4
Write minors and cofactors of the elements of the following determinants:
(i)
(ii)
(i) Let
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) Let
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
Write minors and cofactors of the elements of the following determinants:
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
Using cofactors of elements of second row, evaluate:
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
Using cofactors of elements of third column, evaluate:
If 
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.
Option (D) is correct.
Exercise 4.5
Find adjoint of each of the matrices
Find adjoint of each of the matrices
Verify A (adj A) = (adj A) A = |A| I .
Verify A (adj A) = (adj A) A = |A| I .
Let A =
Find the inverse of the matrix (if it exists) given in Exercise 5 to 11.
Let A =
Let
If A =
For the matrix A =
For the matrix A =
If A =
Let A be a non-singular matrix of order 3 x 3. Then |adjA| is equal to:
(A) |A|
(B) |A|2
(C) |A|3
(D) 3|A|
Therefore, option (B) is correct.
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
Exercise 4.6
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Matrix form of given equations is AX = B
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Examine the consistency of the system of equations
x + 2y = 2
2x + 3y = 3
Matrix form of given equations is AX = B
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
2x − y = 5
x + y = 4
Matrix form of given equations is AX = B
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
x + 3y = 5
2x + 6y = 8
Matrix form of given equations is AX = B
∴ A is a singular matrix.
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Matrix form of given equations is AX = B
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Matrix form of given equations is AX = B
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Solve the system of linear equations, using matrix method
5x + 2y =4
7x + 3y = 5
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
2x – y = – 2
3x + 4y = 3
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
4x – 3y = 3
3x – 5y = 7
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
5x + 2y = 3
3x + 2y = 5
Matrix form of given equations is AX = B
Thus, A is non-singular. Therefore, its inverse exists.
Solve the system of linear equations, using matrix method
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Matrix form of given equations is AX = B
Solve the system of linear equations, using matrix method
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Matrix form of given equations is AX = B
If A = 
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion, 4 kg wheat and 2 kg rice is ` 90. The cost of 6 kg onion, 2 k wheat and 3 kg rice is ` 70. Find cost of each item per kg by matrix method.
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
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.
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