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Byju's Answer
Standard XII
Mathematics
Transpose Conjugate
The value of ...
Question
The value of the determinant
∣
∣ ∣
∣
1
+
i
1
−
i
i
1
+
i
i
1
+
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
A
2
+
5
i
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B
7
−
4
i
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C
4
+
7
i
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D
−
2
+
5
i
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Solution
The correct option is
D
−
2
+
5
i
We have,
∣
∣ ∣
∣
1
+
i
1
−
i
i
1
+
i
i
1
+
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
R
1
→
R
1
−
R
2
R
2
→
R
2
−
R
3
∣
∣ ∣
∣
0
1
−
2
i
−
1
1
−
1
2
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
∴
d
e
t
e
r
m
i
n
a
n
t
=
−
1
[
(
1
−
2
i
)
(
1
−
i
)
−
(
1
+
i
)
(
−
1
)
]
+
i
[
(
1
−
2
i
)
2
i
−
1
]
−
1
−
i
−
(
1
−
2
−
3
i
)
+
i
[
2
i
+
4
−
1
]
=
−
1
−
i
+
1
+
3
i
−
2
+
3
i
=
−
2
+
5
i
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0
Similar questions
Q.
∣
∣ ∣
∣
1
+
i
1
−
i
i
1
−
i
i
1
+
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
=
Q.
∣
∣ ∣
∣
1
+
i
1
−
i
i
1
−
i
i
1
+
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
(where
i
=
√
−
1
) equals
Q.
∣
∣ ∣
∣
1
+
i
1
−
i
1
1
−
i
i
1
+
i
i
1
+
i
1
−
i
∣
∣ ∣
∣
is a
Q.
Without expanding the determinant at any stage,
∣
∣ ∣ ∣
∣
−
5
3
+
5
i
3
2
−
4
i
3
−
5
i
8
4
+
5
i
3
2
+
4
i
4
−
5
i
9
∣
∣ ∣ ∣
∣
,
its value is
Q.
Evaluate:
(
i
)
⎛
⎜ ⎜
⎝
1
−
c
o
t
π
3
1
+
c
o
t
π
3
⎞
⎟ ⎟
⎠
(
i
i
)
1
−
c
o
s
π
6
1
+
c
o
s
π
6
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