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If A is idempotent matrix, then show that `(A+I)^(n) = I+(2^(n)-1) A, AAn in N,` where I is the identity matrix having the same order of A. |
Answer» `because ` A is idempotent matrix `therefore A^(2) = A`, similarly `A = A^(2) = A^(3) =A^(4) =... = A^(n)` …(i) Now, `(A+I)^(n) = (I+A)^(n)` `I+""^(n)C_(1)A+""^(n) C_(2) A^(2)+""^(n)C_(3) A^(3) +...+ ""^(n) C_(n) A^(n) ` `I+(""^(n)C_(1)+""^(n) C_(2) +""^(n)C_(3) +...+ ""^(n) C_(n)) A ` [ from Eq.(i)] `I+(2^(n)-1)A` Hence, `(A+I)^(n) = I + (2^(n)-1) A, AA n in N.` |
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