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Question

lf A is non-singular and (A+I)(A3I)=0 then 3A1A+2I

A
I
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B
0
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C
2I
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D
6I
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Solution

The correct option is B 0
We have,
(A+I)(A3I)=0
A23A+A3I=0[AI=IA=A,I2=I]
A22A3I=0
Multiply both sides by A1
A2I3A1=0[AA1=I]
3A1A+2I=0

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