Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The value of the determinant |1+a2-b2&2ab&-2b 2ab&1-a2+b2&2a 2b&-2a&1-a2-b2| is equal to
Q. The value of the determinant
∣
∣
1
+
a
2
−
b
2
2
ab
2
b
2
ab
1
−
a
2
+
b
2
−
2
a
−
2
b
2
a
1
−
a
2
−
b
2
∣
∣
is equal to
2459
225
WBJEE
WBJEE 2013
Determinants
Report Error
A
0
15%
B
(
1
+
a
2
+
b
2
)
17%
C
(
1
+
a
2
+
b
2
)
2
12%
D
(
1
+
a
2
+
b
2
)
3
57%
Solution:
. Let
Δ
=
∣
∣
1
+
a
2
−
b
2
2
ab
2
b
2
ab
1
−
a
2
+
b
2
−
2
a
−
2
b
2
a
1
−
a
2
−
b
2
∣
∣
Apply
C
1
→
C
1
−
b
C
3
and
C
2
→
a
C
3
+
C
2
Δ
=
∣
∣
1
+
a
2
−
b
2
+
2
b
2
2
ab
−
2
ab
2
b
−
b
+
a
2
b
+
b
3
2
ab
−
2
ab
1
−
a
2
+
b
2
+
2
a
2
−
2
a
+
a
−
a
3
−
a
b
2
−
2
b
2
a
1
−
a
2
−
b
2
∣
∣
=
∣
∣
(
1
+
a
2
+
b
2
)
0
b
(
1
+
a
2
+
b
2
)
0
(
1
+
a
2
+
b
2
)
−
a
(
1
+
a
2
+
b
2
)
−
2
b
2
a
(
1
−
a
2
−
b
2
)
∣
∣
=
(
1
+
a
2
+
b
2
)
2
∣
∣
1
0
b
0
1
−
a
−
2
b
2
a
(
1
−
a
2
−
b
2
)
∣
∣
=
(
1
+
a
2
+
b
2
)
2
{
(
1
−
a
2
−
b
2
+
2
a
2
)
+
2
b
2
}
=
(
1
+
a
2
+
b
2
)
2
(
1
+
a
2
+
b
2
)
=
(
1
+
a
2
+
b
2
)
3