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Tardigrade
Question
Mathematics
| beginmatrix 1 2 3 13 23 33 15 25 35 endmatrix | equal to
Q.
∣
∣
1
1
3
1
5
2
2
3
2
5
3
3
3
3
5
∣
∣
equal to
1978
193
J & K CET
J & K CET 2006
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A
1
!
2
1
3
B
1
!
3
!
5
!
C
6
!
D
9
!
Solution:
Let
A
=
∣
∣
1
1
3
1
5
2
2
3
2
5
3
3
3
3
5
∣
∣
=
1.2.3
∣
∣
1
1
2
1
4
1
2
2
2
4
1
3
2
3
4
∣
∣
=
6
∣
∣
1
1
1
1
4
16
1
9
81
∣
∣
=
6
∣
∣
1
1
1
0
3
15
0
5
65
∣
∣
(
C
2
→
C
2
−
C
1
,
C
3
→
C
3
−
C
2
)
=
6.3.5
∣
∣
1
1
1
0
1
5
0
1
13
∣
∣
=
90
[
1
(
13
−
5
)]
=
90
×
8
=
720
=
6
!