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    				| 1. | If `A`is a square matrix, using mathematical induction prove that `(A^T)^n=(A^n)^T`for all `n in Ndot` | 
| Answer» We will prove it using mathematical induction. For, `n = 1`, `L.H.S = (A^T)^1 = A^T` `R.H.S. = (A^1)^T = A^T` As, `L.H.S.= R.H.S`, our equation is true for `n = 1`. Let, our equation is true for `n = k` where `k` is a natural number. Then, `(A^T)^k = (A^K)^T->(1)`. Now, we have to prove, for `n = k+1`, given equation is true. For, `n = k+1`, `L.H.S. = (A^T)^k+1` `=(A^T)^k*(A^T)^1` From (1), `=(A^k)^T*A^T` `=(A^k*A)^T` `=(A^(k+1))^T` `R.H.S = (A^(k+1))^T`. `:. L.H.S. = R.H.S.`, for `n = k+1`. Thus, our equation is true for `n = k+1`. `:. (A^T)^n = (A^n)^T`. | |