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NDA Math Determinants
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© examsnet.com
Question : 62
Total: 68
If
|
6
i
−
3
i
1
4
3
i
−
1
20
3
i
|
=
x
+
i
y
, where
i
=
√
−
1
, then what is
x
equal to ?
[NDA II 2014]
3
2
1
0
Validate
Solution:
Given,
x
+
i
y
=
|
6
i
−
3
i
1
4
3
i
−
1
20
3
i
|
⇒
x
+
i
y
=
6
i
(
3
i
2
+
3
)
+
3
i
(
4
i
+
20
)
+
1
(
12
−
60
i
)
⇒
x
+
i
y
=
6
i
(
3
i
2
+
3
)
+
12
i
2
+
60
i
+
12
−
60
i
We know that
i
2
=
−
1
⇒
x
+
i
y
=
6
i
[
3
(
−
1
)
+
3
]
+
12
(
−
1
)
+
60
i
+
12
−
60
i
⇒
x
+
i
y
=
0
⇒
x
+
i
y
=
0
+
i
0
⇒
x
=
0
and
y
=
0
Consider,
x
-iy
Put the value of
x
and
y
in above expression, we get
⇒
x
−
i
y
=
0
−
i
0
⇒
x
−
i
y
=
0
© examsnet.com
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