NCERT Solutions for Class 12 Maths Exercise 4.2 Chapter 4 Determinants – FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants is a sure-shot way of obtaining the complete marks in the particular chapter for Board Exam 2019- 2020. CoolGyan provides you with Free PDF download of the same solved by Expert Teachers as per NCERT (CBSE) Book guidelines. They provide the students with precise and to the point answers which fetch very good marks in Board Exams. Download today the NCERT CBSE Solutions for Class 12 Maths Chapter 4 – Determinants to achieve your goal score. Class 12 Maths Chapter 4 – Determinants solved by Expert Teachers as per NCERT (CBSE) Book guidelines.

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

Using the properties of determinants and without expanding in Exercise 1 to 7, prove that:
1. 

Ans. Given:   
Operating 

 0 = 0 [  C₁ and C₂ are identical]
 L.H.S. = R.H.S.

2.   = 0

 

Ans.   
Operating 
= 
=>  = 0 = R.H.S.
[  All entries of one column here first are zero]

3. 

 

Ans.   
operating 
=∣∣∣∣235789637281∣∣∣∣=|276338725981|
= 
= 9 x 0 = 0 [  two columns are identical] Proved.

4. 

 

Ans.   
= 
Operating 

= 
=   = 0 [  two columns are identical] Proved.

5. 

 

Ans. L.H.S. =   
operating 
= 
= 
= 
[operating  ]
= 
[operating  ]
= 
[operating  ]
= 
[Interchanging   and  ]
= 
[Interchanging   and  ]
= 
=   = R.H.S.

6.   = 0

 

Ans. Let   
[Taking   common from each row]
=> 
Interchanging rows and columns in the determinants on R.H.S.,

 

 
   Proved.

7. 

 

Ans. L.H.S. =   
Taking common   from   respectively,
= 
[operating  ]
= 
= 
= 
=   =   = R.H.S.

8. (i) 

 
(ii) 

Ans. (i) L.H.S. =   
 and  ,
= 
∣∣∣∣∣100ab−ac−aa2(b−a)(b+a)(c−a)(c+a)∣∣∣∣∣|1aa20b−a(b−a)(b+a)0c−a(c−a)(c+a)|
Taking (b-a) and (c-a) common from R₂ and R₃ respectively.
=(b−a)(c−a)∣∣∣∣∣100a11a2(b+a)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa201(b+a)01(c+a)|
R2→R2−R3R2→R2−R3
=(b−a)(c−a)∣∣∣∣∣100a01a2(b−c)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa200(b−c)01(c+a)|
[Expanding along 1^st column]
=(b−a)(c−a)∣∣∣01(b+a)(c+a)∣∣∣=(b−a)(c−a)|0(b+a)1(c+a)|
=(b−a)(c−a)(0−(b−c))=(b−a)(c−a)(0−(b−c))
=(b−a)(c−a)(c−b)=(b−a)(c−a)(c−b)
=   = R.H.S. Proved.
(ii) L.H.S. = 
operating   and 
= 
= 
= 
= 
=  
= 
= 
= 
= 
=   = R.H.S.

9. 

 

Ans.   
L.H.S. = 
[Multiplying   by   respectively]
=1xyz∣∣∣∣∣x2y2z2x3y3z3xyzxyzxyz∣∣∣∣∣=1xyz|x2x3xyzy2y3xyzz2z3xyz|
= 
Taking xyz commong from C₃
= 
= 
[operating  ]
= 
= 
= 
= 
= 
= 
= 
= 
= 
= 
= 
=   = R.H.S.

10. (i) 

 
(ii) 

Ans. (i) L.H.S. =   

= 
Taking 5x+4 common from R₁
= 
[operating   and  ]
= 
= 
=   = R.H.S.
(ii) L.H.S. = 

= 
Taking 3y+k common from C₁
= 
[operating   and  ]
= 
= 
=   = 
= R.H.S.  Proved.

11. (i) 

 
(ii) 

Ans. (i) L.H.S. =   

= 
Taking a+b+c common from R₁
= 

= 
= 
= 
=   = R.H.S. Proved.
(ii) L.H.S. = 

= 
Taking 2(x+y+z) common from C₁
= 

= 
= 
= 
=   = R.H.S. Proved.

12. 

 

Ans. L.H.S. =   

=   T
Taking 1+x+x² common from R₁
= 

= 
= 
= 
= 
= 
= 
= 
= 
=   = R.H.S.      Proved.

13. 

 

Ans. L.H.S. =   

= 
= 

= 
= 
= 
=   = R.H.S.

14. 

 

Ans. L.H.S. =   
Multiplying   by   respectively and then dividing the determinant by 
= 
Taking a,b, and c common from R₁, R₂ and R₃ respectively.
= 

= 
= 

= 
= 
=   = R.H.S.        Proved.

Choose the correct answer in Exercises 15 and 16.

 
15. Let A be a square matrix of order 3 x 3, then  is equal to:
(A) 
(B) 
(C) 
(D) 

Ans. Let A =   be a square matrix of order 3 x 3.   ……….(i) 
  
  
  
=   [From eq. (i)]
Therefore, option (C) is correct.