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If `A` is a square matrix such that `A^2=A ,` show that `(I+A)^3=7A+Idot`

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`(I+A)^{3}-7 A=I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A`

`=I+A^{3}+3 A+3 A^{2}-7 {A}`

`=I+A^{2} cdot A+3 A+3 A-7 A quad[A^{2}=A]`

`=I+A cdot A-A`

`=I+A-A`

`=I`

`therefore(I+A)^{3}-7 A=I`

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