NCERT Solutions class 12 Maths Exercise 4.5 Ch 4 Determinants

NCERT Solutions for Class 12 Maths Exercise 4.5 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



Find adjoint of each of the matrices in Exercise 1 and 2.
1. 
Ans. Here A =   
 
 A11 = Cofactor of 
A12 = Cofactor of 
A21 = Cofactor of 
A22 = Cofactor of 
 adj. A =∣∣∣A11A21A12A22∣∣∣=|A11A12A21A22|′
=∣∣∣4231∣∣∣=|4−3−21|′
=∣∣∣4321∣∣∣=|4−2−31|

2. 

 

Ans. Here A =   
 

 = 







 adj. A =∣∣∣∣31111251625∣∣∣∣=|3−126152−11−15|′
=∣∣∣∣31261521115∣∣∣∣=|31−11−125−1625|
 

Verify A (adj. A) = (adj. A) A=    in Exercise 3 and 4.

 
3. 

Ans. Let A =   
 adj. A = 
 A.(adj. A) = 
 =    …..(i)
Again  (adj. A). A = 
 =      …..(ii)
And   = 
Again       …..(iii)
 From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A = 

4. 

 

Ans. Let A =   
   
      = 
  
  
   

 adj. A =∣∣∣∣0321118013∣∣∣∣=|0−11031−1283|′
=∣∣∣∣0110311283∣∣∣∣=|032−11180−13|
 A. (adj. A) = 

 ……….(i)
Again  (adj. A). A = 

  ……….(ii)
And 

Also   =      ……….(iii)
 From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A = 

Find the inverse of the matrix (if it exists) given in Exercise 5 to 11.

 
5. 

Ans. Let A =   
   = 6-(-8) = 6+8 = 14014≠0
 Matrix A is non-singular and hence   exist.
Now adj. A =   And 

6. 

 

Ans. Let A =   
   = 
 Matrix A is non-singular and hence   exist.
Now adj. A =   And 

7. 

 

Ans. Let A =   
   = 
   exists.
A11 =  ,  A12 =  ,
A13 =  ,   A21 =  ,
A22 =  ,   A23 =  ,
A31 =  ,   A32 =  ,
A33 = 
 adj. A =∣∣∣∣10102054002∣∣∣∣=|1000−10502−42|′
=∣∣∣∣10001050242∣∣∣∣=|10−10205−4002|
 

8. 

 

Ans. Let A =   
    = 
   exists.
A11 =  , A12 =  ,
A13 =  , A21 =  ,
A22 =  , A23 =  ,
A31 =  ,   A32 =  ,
A33 = 
 adj. A = 
 

9. 

 

Ans. Let A =   
   = 
   exists.
A11 =  , A12 =  ,
A13 =  , A21 =  ,
A22 =  , A23 =  ,
A31 =  ,  A32 =  ,
A33 = 
 adj. A = 
 

10. 

 

Ans. Let A =   
  
   exists.
A11 =  ,  A12 =  ,
A13 =  , A21 =  ,
A22 =  , A23 =  ,
A31 =  , A32 =  ,
A33 = 
 adj. A = 
 

11. 

 

Ans. Let A =   
 

   exists.
A11 =  ,

A12 =  , A13 =  ,
A21 =  ,   A22 =  ,
A23 =  ,  A31 =  ,
A32 =  , A33 = 
 adj. A = 
 

12. Let A =   and B =   verify that 

 

Ans. Given: Matrix A =   
   = 15 – 14 = 1   0
   = 
Matrix B = 
   = 54 – 56 =     0
 
Now AB =   =   = 
   = 
Now L.H.S. =      ……….(i)
R.H.S. = 

  ……….(ii)
 From eq. (i) and (ii), we get
L.H.S. = R.H.S.
 

13. If A =  , show that A2 – 5A + 7I = 0. Hence find 

 

Ans. Given: A =   
   
 
L.H.S. = 





= R.H.S.
   ……(i)
To find:  , multiplying eq. (i) by  .
 
 
 

 = 

14. For the matrix A =   find numbers a and b such that 

 

Ans. Given: A =   
   
 
 
 
 
 
 We have   ……….(i)
   ……………..(ii)
 
 
Here   satisfies   also, therefore  
Putting   in eq. (i),            
Here also   satisfies   , therefore    
Therefore,   and 

15. For the matrix A =  , show that   Hence find 

 

Ans. Given: A =   
 
 

Now 


L.H.S. = 



 =   = R.H.S.
Now,   to find  , multiplying   by 
 
 
 
 
 
 
=> 11A-1 
 

16. If A =  , verify that   and hence find 

 

Ans. Given: A =   
 
   = 
Now 


L.H.S. = 



 =   = R.H.S.
Now,   to find  , multiplying   by 
 
 
 
 
 
   = 
 

17. Let A be a non-singular matrix of order 3 x 3. Then   is equal to:

 
(A) 
(B) 
(C) 
(D) 

Ans. If A is a non-singular matrix of order   then   
 Putting   
Therefore, option (B) is correct.

18. If A is an invertible matrix of order 2, then det   is equal to:

 
(A) det A
(B) 
(C) 1
(D) 0

Ans. Since   
  
 
 
Therefore, option (B) is correct.

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