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UPSC NDA I 2017 Mathematics Solved Paper
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© examsnet.com
Question : 12
Total: 120
If
A
=
[
α
2
2
α
]
and
d
e
t
(
A
3
)
=
125
, then α is equal to
[NDA I 2017]
±
1
±
2
±
3
±
5
Validate
Solution:
A
=
[
α
2
2
α
]
⇒
A
2
=
[
α
2
+
4
4
α
4
α
α
2
+
4
]
⇒
A
3
=
[
α
3
+
12
α
6
α
2
+
8
6
α
2
+
8
α
3
+
12
α
]
⇒
|
A
3
|
=
125
⇒
(
α
3
+
12
α
)
2
−
(
6
α
2
+
8
)
2
=
125
putting
α
=
±
3
,
(
63
)
2
−
(
62
)
2
=
125
satisfied
© examsnet.com
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