NCERT Solutions for Class 12 Maths Chapter 4 Determinants

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths 4 Determinants

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 4
Chapter Name	Determinants
Exercise	Ex 4.1, Ex 4.2, Ex 4.3, Ex 4.4,Ex 4.5,Ex 4.6
Number of Questions Solved	68
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Chapter 4 Determinants Exercise 4.1

Question 1.
Evaluate the following determinant:

Solution:

= 2x(-1)-(-5)x(4)
=-2+20
=18

Question 2.
(i)
(ii)
Solution:
(i)
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1

Question 3.
If then show that |2A|=|4A|
Solution:

=>
L.H.S = |2A|
=
= – 24

Question 4.
, then show that |3A| = 27|A|
Solution:
3A =
=

Question 5.
Evaluate the following determinant:
(i)
(ii)
(iii)
(iv)
Solution:
(i)

Question 6.
If , find |A|
Solution:
|A| =
= 1(-9+12)-1(-18+15)-2(8-5)
= 0

Question 7.
Find the values of x, if
(i)
(ii)
Solution:
(i)
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii)
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2

Question 8.
If , then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b)
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6

Chapter 4 Determinants Exercise 4.2

Using the property of determinants and without expanding in Q 1 to 5, prove that

Question 1.

Solution:
L.H.S =
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Question 2.

Solution:
L.H.S =

Question 3.

Solution:

Question 4.

Solution:
L.H.S =

Question 5.

Solution:
L.H.S = ∆ =

By using properties of determinants in Q 6 to 14, show that

Question 6.

Solution:
L.H.S = ∆ = …(i)

Question 7.

Solution:
L.H.S =

Question 8.
(a)
(b)
Solution:
(a) L.H.S =

Question 9.

Solution:
Let ∆ =
Applying R1–>R1 – R2, R2–>R2 – R3

Question 10.
(a)
(b)
Solution:
(a) L.H.S =

Question 11.
(a)
(b)
Solution:
(a) L.H.S =
=

Question 12.

Solution:
L.H.S =

Question 13.

Solution:
L.H.S =

Question 14.

Solution:
Let ∆ =

This may be expressed as the sum of 8 determinants

Question 15.
If 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:
Option (c) is correct.

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
Solution:
Option (c) is correct

Chapter 4 Determinants Exercise 4.3

Question 1.
Find the 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) Area of triangle =
= [1(0-3)+1(18-0)]
= 7.5 sq units

Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:
The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)

Question 3.
Find the value 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:
(i) Area of ∆ = 4 (Given)

= [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ==> k = 0
Case(b): -k + 4 = -4 ==> k = 8
Hence, k = 0,8
(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)

Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Given: Points (1,2), (3,6)
Equation of the line is

Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ =
= [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

Chapter 4 Determinants Exercise 4.4

Question 1.
Write the minors and cofactors of the elements of following determinants:
(i)
(ii)
Solution:
(i) Let A =
M₁₁ = 3, M₁₂ = 0, M₂₁ = – 4, M₂₂ = 2
For cofactors

Question 2.
Write Minors and Cofactor of elements of following determinant
(i)
(ii)
Solution:
(i) Minors M₁₁ = = 1

Question 3.
Using cofactors of elements of second row, evaluate

Solution:
Given

Question 4.
Using Cofactors of elements of third column, evaluate

Solution:
Elements of third column are yz, zx, xy

Question 5.
If and Aij is the cofactors of aij? then value of ∆ is given by
(a) a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃
(b) a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁
(c) a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃
(d) a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁
Solution:
Option (d) is correct.

Chapter 3 Determinants Exercise 4.5

Find the adjoint of each of the matrices in Questions 1 and 2.

Question 1.

Solution:
Let C_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
C₁₁ = (-1)¹⁺¹ (4) = 4; C₁₂ = (-1)¹⁺² (3) = -3
C₂₁ = (-1)²⁺¹ (2)= – 2; C₂₂ = (-1)²⁺² (1) = -1
Adj A =
=

Question 2.

Solution:

Similarly,

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Question 3.

Solution:
|A| = 24

Question 4.

Solution:
A₁₁ = 0, A₁₂ = – 11, A₁₃ = 0,
A₂₁ = – 3, A₂₂ = 1, A₂₃ = 1, A₃₁ = – 2
A₃₂ = 8, A₃₃ = – 3

Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:

Question 5.

Solution:

So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Question 6.

Solution:

So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Question 7.

Solution:
|A| = 10

Question 8.

Solution:

So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by

Question 9.

Solution:
|A| =
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.

Question 10.

Solution:
|A| =
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and

Question 11.

Solution:
A =
adj A =
First find |A| = -cos²α-sin²α
=-1≠0

Question 12.
Let , verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| =
= 15-14
= 1≠0

Question 13.
If show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A =
A² =

Question 14.
For the matrix A = find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A =
A²+aA+bI²=0

Question 15.
For the matrix  Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² =

Question 16.
If show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have

Question 17.
Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:
Let A =


Dividing by | A |, |Adj. A| = | A |²
Hence, Part (b) is the correct answer.

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
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I

Hence option (b) is correct.

Chapter 4 Determinants Exercise 4.6

Examine the consistency of the system of equations in Questions 1 to 6:

Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=>
=> AX = B
Now |A| =
= 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.

Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=>
=> AX = B
Now |A| =
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.

Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=>
=> AX = B
Now |A| =
= 6 – 6
= 0.

Hence, equations are consistent with no solution

Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z =

Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:

=> AX = B

Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1


= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.

Solve system of linear equations using matrix method in Questions 7 to 14:

Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as

Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written

Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as

where

Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as

where

Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A^-1B

Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written

Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:

Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written

Question 15.
If A = Find A^-1. Using A^-1. Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
We have AX = B
where

Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. 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 the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70

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