InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If `A=[[1, 1, 1],[ 1, 1, 1],[ 1, 1, 1]]`, then prove that `A^n=[[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)]]`for every positive integer `n`. |
|
Answer» We shall prove the result by using the principle of mathematical induction. When n=1, we have `A^(1)=[{:(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)):}]=[{:(3^(0),3^(0),3^(0)),(3^(0),3^(0),3^(0)),(3^(0),3^(0),3^(0)):}]=[{:(1,1,1),(1,1,1),(1,1,1):}].` Thus, the result is true for `n=1`. Let it be true for n=k. Then, `A^(k)=[{:(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)):}].` `:." "A^(k-1)=A.A^(k)` `=[{:(1,1,1),(1,1,1),(1,1,1):}][{:(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)):}]` `=[{:(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))),(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))),(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))):}]=[{:(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)):}].` Thus, the result is true for `n=(k+1)`, whenever it is true for `n=k.` So, the result is true for all `n in N.` Hence, `A^(n)=[{:(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)):}]`, for all values of `n inN.` |
|