If A is a square matrix such that A^2=I,then find the simplified value of (A-I)^3+(A+I)^3-7A.

If A^2 is i then A^3 is also i now expand the brackets with cube and square fourmula then put the values
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=a is the required answer.
 
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on expanding,
=A^3 - i^3 - 3A^2 + 3A + A^3 + I^3 + 3A^2 + 3A -7A
=2A^3 +6A-7A
=2A^3-A
as we know that A^2=I
   A^3=^2 *A=IA=A
 the simplified form = 2A-A=A
thus A is the required answer.

 
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= a
 
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how a its coming 8A-7I
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Answer is A as after simplying the ques it comes to be 8A -7A
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(a) We have, A2 = I
Also A and I are commutative, so we can expand (A+I)n using expansion of (a+b)n, where a and b ∈ C
∴ (A-I)3 + (A+I)3 - 7A
= A3 - 3A2 + 3A - I3 + A3 + 3A2 + 3A + I3 - 7A
= 2A3 + 6A - 7A
= 2A2 . A + 6A - 7A
= 2I.A + 6A - 7A
= 2A + 6A - 7A = 8A - 7A = A 
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Apply (a-b)^3 and (a+b)^3 so,

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Answer is A or if you are using I then it is I
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We have, A2 = I Also A and I are commutative, so we can expand (A+I)n using expansion of (a+b)n, where a and b ∈ C ∴ (A-I)3 + (A+I)3 - 7A = A3 - 3A2 + 3A - I3 + A3 + 3A2 + 3A + I3 - 7A = 2A3 + 6A - 7A = 2A2 . A + 6A - 7A = 2I.A + 6A - 7A = 2A + 6A - 7A = 8A - 7A = A
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