Mathematics Times – July 2019

vanish if the two zero entries are in the same

row or the same column because then two

columns (or rows) of A would be equal. In all

other cases, by a permutation of rows and

columns the matrix A can be converted to the

matrix

0 1 1

1 0 1

1 1 1

B

 

 

 

 

. But then det (B) = 1

and so det ( )A   1 0. So the only matrices

in E 1 are those where the two zero entries

are in the same row or in the same column.

These two possibilites are mutually exclusive.

In each possibility the exceptional row or

column ( the one containing both the zeros)

can be chosen in 3 ways, and for each such

choice the two zeros can be put into the three

possible places in 3 ways. Putting it all together,

E E 1       2 3 3 3 3 18 (4)

So 1 2

18

( / ) 0.

36

P E E  

15.Sol: Note that  and ^2 are complex

conjugates of each other. Also, ^3  1. So,

22 2

a b c     (a b c )

(^) (a b c  ^2 )
  (a b c a b c ^2 )(   ^2 )
     a b c ab bc ca^222 ( )( ^2  ) (1)
     a b c ab bc ca^222 (2)
(^1) ( ) (^2) ( ) (^2) ( ) 2
2
 a b b c c a      (3)
where in going from (1) to (2) we have used
that  ^2   1 0 which follows from the
factorisation 0 (   ^3 1) (  1)(^2  1)
because the factor  1 is non-zero.
As a b c, , are distinct non-zero integers, the
bracketed expression in (3) will be minimum
when two of the differences a b b c , and
c a are 1 and the third one is 2. So the
minimum value of the bracketed expression is
1 1 4 6  . Hence the minimum of the given
expression is 3.
16.Sol:
Distance of point A from given line
5
2

2
1
CA
CB


2
1
AC
AB

^ AC  2 5 10
17.Sol: We equate the general terms of three
respective Arithmetic progression as
i.e., 1 3    a 2 5b 3 7c
 3 divides 1 + 5b and 5 divides 1+2c
i.e., 1+2c is multiple of 5
So, first such terms are possible when
1 2 c 15 i.e., c 7
Hence, first term a 52 and common
difference d1cm (3,5,7)=
 a d  157
18.Sol: Given that
15.Sol:
16.Sol:
17.Sol:
18.Sol:
/ 4
sin
/ 4
2
(1 x)(2 cos 2 )
dx
I
e x




   ...(1)

Using a b x  properly, we have

/ 4 sin

sin

4

2

(1 )(2 cos 2 )

x

x

e dx

I dx

e x







   ...(2)

adding (1) and (2), we get

4 sin

sin

4

2 (1 )

2

(1 )(2 cos 2 )

x

x

e

I dx

e x









  