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Byju's Answer
Standard XII
Mathematics
Definition of a Determinant
Prove that ...
Question
Prove that
∣
∣ ∣ ∣
∣
1
ω
ω
2
ω
ω
2
1
ω
2
1
ω
∣
∣ ∣ ∣
∣
=
0
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Solution
Solving given determinant, we get
[
1
(
ω
3
−
1
)
−
ω
(
ω
2
−
ω
2
)
+
ω
2
(
ω
−
ω
4
)
]
....... as
(
ω
3
=
1
)
⟹
[
(
1
−
1
)
−
ω
(
0
)
+
ω
2
(
ω
−
ω
)
]
⟹
0
+
0
+
0
⟹
0
Hence proved
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Similar questions
Q.
Show with out expanding
∣
∣ ∣ ∣
∣
1
ω
ω
2
ω
ω
2
1
ω
2
1
ω
∣
∣ ∣ ∣
∣
=
0
. Where
ω
being cube root of unity.
Q.
If
ω
is the cube root of unity, then
∣
∣ ∣ ∣
∣
1
ω
ω
2
ω
ω
2
1
ω
2
1
ω
∣
∣ ∣ ∣
∣
=
Q.
If
ω
=
−
1
2
+
√
3
i
2
, the value of the determinant
∣
∣ ∣ ∣
∣
1
ω
ω
2
ω
ω
2
1
ω
2
1
ω
∣
∣ ∣ ∣
∣
is
Q.
Prove that 0/0 =2
Q.
If
ω
is one of the imaginary cube roots of unity, find the value of
∣
∣ ∣ ∣
∣
1
ω
ω
2
ω
ω
2
1
ω
2
1
ω
∣
∣ ∣ ∣
∣
.
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