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Q.

If A is non-singular and A2–5A+7I=0 then I=

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a

17A−57A−1

b

17A+57A−1

c

15A+75A−1

d

15A−A−1

answer is C.

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Detailed Solution

A2−5A+7I=0 By right multiplication with A-1AAA-1-5AA-1+7IA-1=0 A-5I+7A-1=0⇒ 7A−1 =5I−A ⇒ I=75 A−1+15A
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