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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.