Mathematics Times – July 2019

For non-trivial solution, we have

i.e.,

1 1 1

1 1 1 0

1 1 1

a

b

c

  

   

  

R R R 2   2 3

1 1 1

0 0

1 1 1

a

b c

c

  

 

  

C C C 2   2 3

1 0 1

0 0

1 1

a

b c c

c c

 

  

  

Apply R R R 3   3 1

1 0 1

0 0

a

b c c

a c c

 

  

 

 a bc c c 1   ^22  1 a b c   0

abc bc ab ac    0

   ab bc ca abc

18.Sol: Given  

11 12

11 22

21 22

, ij 0,1,2 ,

a a

a a a

a a

 

   

 

Now, Det of the given matrix is

 a a a a11 22 12 21

Also given   0

 The possible det values are

   4, 3, 2, 1,1,2,3,4

 There are 20 non singular matrices

19.Sol: Let

a b

A

c d

 

 

 

given A I^2 

i.e.,

2

2

1 0

0 1

a bc ab bd

ac cd bc d

     

  

     

 a bc c ab cd^2   1;  0 ;

ac cd bc d 0 ;  ^21

Solving above equations, we get a d 

 a d  0

 Statement-1 is correct

Now detA ad bc 

Now a bc bc d^2     1 2

 a d^2 ^2

Now, we have a bc bc d^2    ^22

 a d ad bc ^2  2  2  2

i.e.,ad bc  1

 det A 1 

So statement-2 is true.

20. Sol:

a b c a b c a b c a b c

b c a b c a

c a b c a b

     



 

1 1 1

a b c b c a

c a b

  

 

18.Sol:

19.Sol:

20. Sol:

21.Sol:

0 0 1

a b c b c c a a

c a a b b

    

 

  a b c ab bc ca a b c     ^222 

       

   a b c a b b c c a ^2  ^2  ^2 

 

Since a,b,c are sides of a scalene triangle, therefore

at least two of the a,b,c will be unequal.

 a b b c c a ^2    ^2   ^2  0

Also a b c   0

       

   a b c a b b c c a ^2  ^2  ^2  0

 

21.Sol: Given system of equations is homogeneous

which isx ay  0