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Let A and B are two matrices such that `AB = BA,` then for every `n in N`A. `A^(n) B = BA^(n)`B. `(AB)^(n) = A^(n)B^(n)`C. `(A+B)^(n) = ""^(n)C_(0) A^(n) + ""^(n)C_(1) A^(n-1) B+...+ ""^(n) C_(n) B^(n) `D. `A^(2n) - B ^(2n) = (A^(n)-B^(n) ) (A^(n)+B^(n))` |
Answer» Correct Answer - A::C::D `because A^(2) B = A (AB) = A(BA) = (AB)A = (BA)A= BA^(2)` Similarly, `A^(3) B = BA^(3) ` In general, `A^(n) B = BA^(n) , AA n ge 1 ` and `(A +B) ^(n) = ""^(n) C_(0)A^(n) + ""^(n)C_(1)A^(n-1) B +""^(n)C_(2) A^(n-2) B^(2) +...+""^(n) C_(n) B^(n)` Also, `(A^(n) - B^(n)) (A^(n) + B^(n) ) = A^(n) A^(n) + A^(n) B^(n) - B^(n) A^(n) - B^(n) B^(n) ` `= A^(2n)-B^(2n) [ because AB = BA]` |
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