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Let A be a square matrix with real entries such that (A^2)=A.Then

a)((A+I))^10=I+1024A

b)((A+I))^10=I+511A

c)((A+I))^10=I+1023A

d)((A+I))^10=I+512A

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Given A^2=A i.e., A is an Idempotent Matrix.

To find : (A+I)^10

If we calculate recursively uptil (A+I)^4 we will arrive at a pattern i.e.,

(A+I)^4=9A^2+6A+I= 15A+I (since A^2=A)

In general, (A+I)^n=(2^n – 1)A + I, i.e., (A+I)^10=I + 1023A , which is option c).

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