If `A=[{:(1,0,0),(1,0,1),(0,1,0):}]`, then which is true (a) `A^(3)-A^(2)=A-I` (b) det. `(A^(100)-I)=0` (c) `A^(200)=[(1,0,0),(100,1,0),(100,0,1)]` (d) `A^(100)=[(1,1,0),(50,1,0),(50,0,1)]`

Options

A. `A^(3)-A^(2)=A-I`
B. det. `(A^(100)-I)=0`
C. `A^(200)=[(1,0,0),(100,1,0),(100,0,1)]`
D. `A^(100)=[(1,1,0),(50,1,0),(50,0,1)]`

Explanation

`A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]`
`A^(3)=[(1,0,0),(1,0,1),(0,1,1)][(1,0,0),(1,1,0),(1,0,1)]=[(1,0,0),(2,0,1),(1,1,0)]`
`A^(3)=A^(2)=[(0,0,0),(1,-1,1),(0,1,-1)]`
and `A-I = [(0,0,0),(1,-1,1),(0,1,-1)]` ltvbrgt `implies A^(3)-A^(2)=A-I` (1)
Now, det `(A^(n)-I)=` det. `((A-I)(I+A+A^(2)+...+A^(n-1)))`
`=" det."(A-1)xx"det. "(I+A+A^(2)+...+A^(n-1))`
`=0`
From (1), `A^(4)-A^(3)=A^(2)-A` (2)
Again from (1), `A^(5)-A^(4)=A^(3)-A^(2)=A-I`
Thus, if `n` is even, `A^(n)-A^(n-1)=A^(2)-A` (3)
If `n` is odd, `A^(n)-A^(n-1)=A-I` (4)
Consider the n is even.
`:. A^(n)-A^(n-1)=A^(2)-A` [from (3)]
and `A^(n-1)-A^(n-2)=A-I` [from (4)]
Adding these, we get
`A^(n)-A^(n-2)=A^(2)-I`
`implies A^(n)=A^(n-1)+A^(2)-I`
`=(A^(n-1)+A^(2)-I)+A^(2)-I`
`=(A^(n-6)+A^(2)-I)+2(A^(2)-I)`
`{:(...,...,...),(...,...,...):}`
`=(A^(2))+(n-2)/2 (A^(2)-I)`
`:. A^(n)=(n/2) A^(2)-((n-2)/2)I`
`:. A^(200)=100 A^(2)-991`
`=100 [(1,0,0),(1,1,0),(1,0,1)]-99 [(1,0,0),(0,1,0),(0,0,1)]=[(1,0,0),(100,1,0),(100,0,1)]`