UPSC NDA 2 2017 Mathematics Solved Paper

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Question : 11 of 120

Marks: +1, -0

If A=[

4i−6	10i
14i	6−4i

] and k=

1

2i

where i=√−1 then kA is equal to

[NDA II 2017]

[
2+3i 5
7 2−3i
]
[
2−3i 5
7 2+3i
]
[
2−3i 7
5 2+3i
]
[
2+3i 5
7 2+3i
]

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