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NCERT Solutions class 12 Maths Exercise 3.4 Ch 3 Matrices

NCERT Solutions for Class 12 Maths Exercise 3.4 NCERT Solutions for Class 12 Maths Chapter 3 – Matrices – FREE PDF Download

NCERT Class 12 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 12 Maths Chapter 3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. Class 12 Maths Chapter 3 – Matrices solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Matrices Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 3 – Matrices



Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.

1.  

Ans. Let A =  

Since A = IA     

    

    

    

   = 


2. 

 

Ans. Let A =  


3. 

 

Ans. Let A =  

Since A = IA     

      

     

   = =>[7231]=>[7−3−21]


4. 

 

Ans. Let A =  

Since A = IA     

      

=>[1213]=[2110]A=>[1−123]=[−2−110]A 

    [R2R22R1][R2→R2−2R1]

    

   = 


5. 

 

Ans. Let A =  

Since A = IA     

      

=>[1101]=[4311]A=>[1011]=[4−1−31]A 

=>[1001]=[4712]A=>[1001]=[4−1−72]A   

   = [4712][4−1−72]


6. 

 

Ans. Let A =  

Since A = IA     

=>[1235]=[0110]A=>[1325]=[0110]A    

     

      

     

   = 


Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 7 to 14.

 

7. 

Ans. Let A =  

Since A = IA     

      

     

   [R2R25R1][R2→R2−5R1]

     

   = 


8. 

 

Ans. Let A =  

Since A = IA     

      

     

     

   = 


9. 

 

Ans. Let A =  

Since A = IA     

      

     

     

   = 


10. 

 

Ans. Let A =  

Since A = IA     

      

    [R1(1)R1][R1→(−1)R1]

  =>[1012]=[1413]A=>[1−10−2]=[−1−1−43]A   

     

     

   = 


11. 

 

Ans. Let A =  

Since A = IA     

  =>[1226]=[0110]A=>[1−22−6]=[0110]A   

     

     

     

   = 


12. 

 

Ans. Let A =  

Since A = IA     

     

     

Here, all entries in second row of left side are zero.

   does not exist.


13. 

 

Ans. Let A =  

Since A = IA     

     

      

      

   = 


14. 

 

Ans. Let A =  

Since A = IA     

     

     

Here, all entries in second row of left side are zero.

   does not exist.


 

15. 

Ans. Let A = 

We know that A = IA,

  

    

    

    

    

    

    

   

   [R2R2+R3][R2→R2+R3]

  


16. 

 

Ans. Let A = 

Since, A = IA

  

    

    

    

    

    

=>⎡⎣⎢⎢100010001⎤⎦⎥⎥=⎡⎣⎢⎢⎢⎢1253525425125351125925⎤⎦⎥⎥⎥⎥A=>[100010001]=[1−25−35−254251125−35125925]A  

   = 


17. 

 

Ans. Let A = 

Since, A = IA

  

    

    

=>⎡⎣⎢⎢100121253⎤⎦⎥⎥=⎡⎣⎢⎢250120001⎤⎦⎥⎥A=>[1120−2−5013]=[−2105−20001]A  

=>⎡⎣⎢⎢100112235⎤⎦⎥⎥=⎡⎣⎢⎢205102010⎤⎦⎥⎥A=>[1120130−2−5]=[−2100015−20]A  

    

    

   = 


18.  Matrices A and B will be inverse of each other only if:

 

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans. By definition of inverse of square matrix, 

Option (D) is correct.