TS ICET 23 May 2019 Shift 1 Paper

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Question : 132 of 200

Marks: +1, -0

A =

[\table 1,1;1,1]

⇒

A^{2019}

=

$2^{2020} A$
$2^{2019} A$
$2^{2018} A$
$2^{2017} A$

Solution:

A=[\table 1 , 1 ; 1 , 1]

A^{2}=[\table 1 , 1 ; 1 , 1][\table 1 , 1 ; 1 , 1]=.[\table 2 , 2 ; 2 , 2]

A^{4}=[\table 2 , 2 ; 2 , 2][\table 2 , 2 ; 2 , 2]=[\table 8 , 8 ; 8 , 8]

A^{4}=2^{3}[\table 1 , 1 ; 1 , 1]=2^{3} A

Similarly

A^5=2^4A

∴A^{2019}=2^2018A

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