AP ICET 2016 Paper

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Question : 139 of 200

Marks: +1, -0

If A is an n x n matrix such that

A^2=I

and

A≠A^2

,then

{\det}(I+A)=

0
1
$-1$
2

Solution:

Given : A is an n × n matrix

\A^2=\I

and

\A≠\A^2

Let

\A=(\table-3,4;-2,3)

\A^2=(\table-3,4;-2,3)(\table-3,4;-2,3)(\table9-8,-12+12;6-6,-8+9)

=(\table1,0;0,1)=\I

Also,

\A≠\A^2

\I+\A=(\table1,0;0,1)+(\table-3,4;-2,3)=(\table-2,4;-2,4)

\det(\I+\A)=[\table-2,4;-2,4]=(–8)–(–8)=0

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