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Byju's Answer
Standard XII
Mathematics
Determinant
Find a if ...
Question
Find
a
if
∣
∣ ∣
∣
i
−
2
i
−
1
3
i
i
3
−
2
1
−
3
−
i
∣
∣ ∣
∣
=
a
i
, where
i
=
√
−
1
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Solution
Given that
∣
∣ ∣
∣
i
−
2
i
−
1
3
i
i
3
−
2
1
−
3
−
i
∣
∣ ∣
∣
=
a
i
Expanding given Determinant
i
(
−
i
4
−
6
)
+
2
i
(
−
3
i
2
+
2
)
−
1
(
−
9
i
−
i
3
)
=
−
i
5
−
6
i
−
6
i
3
+
4
i
+
9
i
+
i
3
=
−
i
−
6
i
+
6
i
+
4
i
+
9
i
−
i
=
11
i
Henece value of
a
=
11
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0
Similar questions
Q.
If
z
=
−
2
(
1
+
2
i
)
3
+
i
where
i
=
√
−
1
, then the argument
θ
(
−
π
<
θ
≤
π
)
of
z
is
Q.
Find the value of the following determinants, where
i
=
√
−
1
.
(i)
∣
∣
∣
2
i
−
3
i
i
3
−
2
i
5
∣
∣
∣
(ii)
∣
∣
∣
1
+
3
i
i
−
2
−
i
−
2
1
−
3
i
∣
∣
∣
Q.
If
∑
n
i
=
1
→
a
i
=
0
where
∣
∣
→
a
i
∣
∣
=
1
for all
i
then the value of
∑
1
≤
i
<
j
≤
n
∑
→
a
i
→
a
j
is
Q.
If
A
=
⎡
⎢ ⎢ ⎢
⎣
−
1
+
i
√
3
2
i
−
1
−
i
√
3
2
i
1
+
i
√
3
2
i
1
−
i
√
3
2
i
⎤
⎥ ⎥ ⎥
⎦
,
i
=
√
−
i
and
f
(
x
)
=
x
2
+
2
, then
f
(
A
)
is equal to
Q.
If
n
∑
i
=
1
→
a
i
=
0
where
∣
∣
→
a
i
∣
∣
=
1
,
∀
i
,
then the value of
∑
1
≤
i
<
j
≤
n
∑
→
a
i
.
→
a
j
is equal to
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