Mathematics Times – July 2019

Now

   

   

  

1

1

2

2 1 3 2

1

2

1 1 3 4

1 1

2 2

n

r r

r r r

n

n a

n n

n n n





    

  

 

 



We know

 

   

(^112)
1 1
1
; 2 1 1 ;
2
n n
r r
n n
r r n
 
 


     

 

(^1)   
1
1 3 4
3 2
2
n
r
n n
r


 

  

 

 

  

 

 

  

2

1

1

2

1 1 3 4

1

2 2

1

2

1 1 3 4

1

2 2

n

r

r

n n n n

n

n

n a

n n n n

n





  



   

  





(^)  0 R R 1  2 
Hence, value of
1
1
n
r
r



 is independent of both ‘a’

and ‘n’

15. Sol: Given

     

     

(^222222)
2 2 2
(^222) 1 1 1
a b c a b c
a b c k a b c
a b c
   
  
   
  
Now, we take L.H.S      
     
2 2 2
2 2 2
2 2 2
a b c
a b c
a b c
  
  
  
  
R R R 2   3 2
     
2 2 2
2 2 2
4 4 4
a b c
a b c
a b c
  
  

  
     
2 2 2
2 2 2
4
a b c
a b c
a b c
  
  

  
R R R R 3   3  1  2 2 

15. Sol:

16.Sol:

17.Sol:

2 2 2

2 2 2

4

a b c

a b c  

  



2 2 2

4 3

1 1 1

a b c

  a b c

2 2 2

1 1 1

a b c

k a b c given

 k 4 ^2

16.Sol: Given

1 2 3 0 0 1

0 2 3 1 0 0

0 1 1 0 1 0

A

   

   

   

    

Applying C C 1  3

3 2 1 1 0 0

3 2 0 0 0 1

1 1 0 0 1 0

A

   

   

   

    

Again Applying C C 2  3

3 1 2 1 0 0

3 0 2 0 1 0

1 0 1 0 0 1

A

   

   

   

    

pre-multiplying both sides by A^1

1 1

3 1 2 1 0 0

3 0 2 0 1 0

1 0 1 0 0 1

A A A

   

   

   

   

1 1

3 1 2

3 0 2

1 0 1

I A I A 

 

  

 

 

A A I I^1  and identity matrix 

Hence,

1

3 1 2

3 0 2

1 0 1

A

 

 

 

 

17.Sol: Given system of equations can be written as

a x y z^1    ^0

  x b y z (^1)    0
   x y c z 1 0 