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Question

A=⎡⎢⎣123123−1−2−3⎤⎥⎦, then A is a nilpotent matrix of index

A
2
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B
3
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C
4
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D
5
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Solution

The correct option is A 2
Given A=123123123

For A to be nilpotent matrix then Am=0
where m is the index, so let's check for m=2

A2=123123123× 123123123

A=1+232+463+691+232+463+6912+324+636+9

A=000000000

For m=2 ,A is a Nilpotent matrix

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