TS ICET 24 May 2019 Shift 1 Paper

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Question : 133 of 200

Marks: +1, -0

A =

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

, B =

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

⇒

A^{99}B^{100}+B^{99}A^{100}

=

A + B
A - B
AB
0

Solution:

B=[\table 0 , 0 , 1 ; 0 , 1 , 0 ; 1 , 0 , 0]

B ^{2}=[\table 0 , 0 , 1 ; 0 , 1 , 0 ; 1 , 0 , 0][\table 0 , 0 , 1 ; 0 , 1 , 0 ; 1 , 0 , 0]=[\table 1 , 0 , 0 ; 0 , 1 , 0 ; 0 , 0 , 1]= A

∴A^99=A;B^100=A

B^99 = B; A^100 = A

and AB = B

A^99 B ^100 + B^99 A^100

A A + B A = A + B

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