Mathematics Times – July 2019

^2 

51 63

3 12

84 72

adj A A

 

  

 

5.Sol: Given A adjA AA  T

A A adjA A AA^1   ^1 T

adjA A T

2 5 3

3 5 2

b a

a b

   

   

   

(^2) and 3
5
 a b
   5 a b 5

6. Sol: For trivial solution, we have

1 1

1 1 0

1 1









  



    (^1)  1 0
    0, 1, 1
7.Sol:
1 2 3 1
1 2 3 2
1 2 3
2 2
2 3 2
2
x x x x
x x x x
x x x



   
   

   

rewrite the given equation as
 2 x x x 1  22   3 0
2 x 1   3 x x 2 2 0 3 
 x x x 1 22  3  0
For non-trivial solution, we have
  0
i.e.,  
2 2 1
2 3 2 0
1 2



 
  
 
  2    3      4 2 2 2   
1 4 3    0
     ^32 5 3 0
  1,1, 3
Hence  has 2 values.
8.Sol: Given f n   n n
 f (^1)   , 2f ^2 ^2 , 3f ^3 ^3 ,
and f 4   ^44
Let
  
     
     
3 1 1 1 2
1 1 1 2 1 3
1 2 1 3 1 4
f f
f f f
f f f
 
    
  
i.e.,
2 2
2 2 3 3
2 2 3 3 4 4
3 1 1
1 1 1
1 1 1
   
     
     
   
       
     
5.Sol:

6. Sol:

7.Sol:

8.Sol:

9.Sol:

10. Sol:

2

2 2 2 2 2 2

1 1 1 1 1 1 1 1 1

1 1 1

1 1 1

     

     

 

On expanding, we get

     

   1 ^21   ^2 ^2

 k 1  ^21    ^2   ^2   1  ^21    ^2  ^2

 k 1

9.Sol: For non-trivial solution of the given system of

linear equations

4 2

4 1 0

2 2 1

k

k 

 8 (2 ) 2(2 8) 0k k  k 

    k k^2 6 8 0

 k2,4

Clearly there exists two values of k.

10. Sol:

1 8 (^2) 4 3 8
3
k
k k k
k k

     

(^)   k k^2 4 3
 ( 3)( 1)k k
2
1
4 8
4 12 24 8
3 1 3
k
k k k
k k
     
 
(^) 4 12 8k k^2  
4( 3 2)k k^2  
4( 2)( 1)k k 