NIMCET 2018 Question Paper with solutions

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Question : 11 of 120

Marks: +1, -0

If A =

[\table 0,5;0,0]

and f (x) = 1 + x +

x^2

+ ..... +

x^{16}

, then f (A) =

0
$[\table 1,5;0,1]$
$[\table 1,5;0,0]$
$[\table 0,5;1,1]$

Solution:

∴ f (x) = 1 A +

A^2

+ .... +

A^{16}

[put x = A]
Now ,

A^2

= A . A
=

[\table 0,5;0,0][\table0,5;0,0]

=

[\table 0+0,0+0;0+0,0+0]

=

[\table 0,0;0,0]

∴

A^2

=

A^3

= ..... =

A^{16}

= 0
Hence f (A) = I + A
=

[\table 1,0;0,1]

+

[\table 0,5;0,0]

[|I| = 1]
=

[\table 1+0,0+5;0+0,1+0]

=

[\table 1,5;0,1]

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