y az 0
z ax 0
1 0
0 1
0 1
a
A a
a
Now, | | [1 ( )] 1A a a^2 a^30
So, system has only trivial solution.
Now, | | 0A only when a 1
So, system of equations has infinitely many
solutions which is not possible because it is given
that system has a unique solution. Hence set of all
real values of ‘a’ is R { 1}
- Sol: 1
1 sin cos
1 cos sin
1 sin cos
R R R R R R 1 2 1 ; 2 3 2
0 sin cos cos sin
0 cos sin sin cos
1 sin cos
sin cos ^2 cos^2 sin^2
2sin^2 2sin .cos
2sin sin cos
Now, 4 in 0, 2
1 2 sin 0 0,
Since value of sin is finite for 0, 2
Hence non-trivial solution for only one value of
in 0, 2
cos sin cos
sin cos sin 0
cos sin cos
C C C 1 2 1
0 sin cos
0 cos sin 0
2cos sin cos
2cos sin^2 cos^2 ^0
cos 0 or sin^2 cos^2 0
But cos 0 not possible for any value of
0,
2
sin^2 cos^2 0 sin cos , which
is also not possible for any value of 0, 2
Hence, there is no solution.
23.Sol: Given system of equations can be written in
matrix form as AX B where
1 2 3 6
1 3 5 and 9
2 5
A B
a b
Since, system is consistent and has infinitely many
solutions
3 25 15 2 1 6 0
10 6 2 9 0
1 1 1 0
a a
a a
b
6 9 b 0 b 15
and 6 10 a a 9 6 2 b 0
60 6 9 54 30 0a a
3 24a a 8
Hence, a b8, 15.
[OFFLINE QUESTIONS]
1.Sol: now,
- Sol:
23.Sol:
[OFFLINE QUESTIONS]
1.Sol:
2
4 2 2
2 4 2 ( )( )
2 2 4
x x x
x x x A Bx x A
x x x
Put
3
4 0 0
0 0 4 0 4
0 0 4
x A A
