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Question

If A=101012004 and |3A|=k|A|, then k is

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Solution

A=101012004 |A|=4
3A=3030360012
Expanding along the first row
|3A|=336012006012+30300
=3(3×126×0)0(0×120×6)+(0×00×3)
=3(360)+0+3(0+0)
=3×36
=108
=27×4
=27|A|
Hence k=27

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