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NCERT Solutions class 12 Maths Exercise 4.2 Ch 4 Determinants

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 ,

∣∣∣∣∣100abacaa2(ba)(b+a)(ca)(c+a)∣∣∣∣∣|1aa20b−a(b−a)(b+a)0c−a(c−a)(c+a)|

Taking (b-a) and (c-a) common from R2 and Rrespectively.

=(ba)(ca)∣∣∣∣∣100a11a2(b+a)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa201(b+a)01(c+a)|

R2R2R3R2→R2−R3

=(ba)(ca)∣∣∣∣∣100a01a2(bc)(c+a)∣∣∣∣∣=(b−a)(c−a)|1aa200(b−c)01(c+a)|

[Expanding along 1st column]

=(ba)(ca)∣∣∣01(b+a)(c+a)∣∣∣=(b−a)(c−a)|0(b+a)1(c+a)|

=(ba)(ca)(0(bc))=(b−a)(c−a)(0−(b−c))

=(ba)(ca)(cb)=(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, Rand 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.


16. Which is 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.

Ans. Since, Determinant is a number associated to a square matrix. 

Therefore, option (C) is correct.