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Solutions ⇒ Class 12th ⇒ Mathematics-I ⇒ 3. Matrices

Solutions 3. Matrices - Exercise 3.4 | Class 12 Mathematics-I - Toppers Study

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3. Matrices

| Exercise 3.4 |

Solutions 3. Matrices - Exercise 3.4 | Class 12 Mathematics-I - Toppers Study


Execise - 3.4

Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 1 to 17.

Ques.1.  

Ans. Let A = 

Since A = IA     

    

    

    

   = 

 

 

Ques.2.    

Ans. Let A = 

Since A = IA     

     

     

    

   = 

 

 

Ques.3.  

Ans. Let A = 

Since A = IA     

      

     

    

   = 

 

 

Ques.4.      

Ans. Let A = 

Since A = IA     

      

     

    

    

   = 

 

 

Ques.5.    

Ans. Let A = 

Since A = IA     

      

     

     

   = 

 

 

Ques.6.  

Ans. Let A = 

Since A = IA     

      

     

      

     

   = 

 

 

Ques.7.    

Ans. Let A = 

Since A = IA     

      

     

     

     

   = 

 

 

Ques.8.   

Ans. Let A = 

Since A = IA     

      

     

     

   = 

 

 

Ques.9.   

Ans. Let A = 

Since A = IA     

      

     

     

   = 

 

 

Ques.10.  

Ans. Let A = 

Since A = IA     

      

     

     

     

     

   = 

 

 

Ques.11.   

Ans. Let A = 

Since A = IA     

     

     

     

     

   = 

 

 

Ques.12.   

Ans. Let A = 

Since A = IA     

     

     

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

   does not exist.

 

 

Ques.13. 

Ans. Let A = 

Since A = IA     

     

      

      

   = 

 

 

Ques.14.  

Ans. Let A = 

Since A = IA     

     

     

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

   does not exist.

 

Ques.15.     

Ans. Let A = , We know that A = IA,    

    

    

    

    

    

    

   

    

  

 

 

Ques.16.    

Ans. Let A = , Since, A = IA     

    

    

    

    

    

    

   = 

 

 

Ques.17.  

Ans. Let A = , Since, A = IA     

    

    

    

    

    

    

   = 

 

 

Ques.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 (A) is correct.

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