What is the inverse of `A=[{:(1+i,1+i),(-1+i,1-i):}]`?

Options

A. `(1)/(4)[{:(1-i,-1-i),(1-i,1+i):}]`
B. `(1)/(4)[{:(1+i,-1+i),(1+i,-1-i):}]`
C. `(1)/(4)[{:(1+i,1-i),(-1-i,1+i):}]`
D. `(1)/(4)[{:(1+i,1-i),(-1-i,-1+i):}]`

Answer

A

Explanation

We know that
`If A = [{:(a,b),(c,d):}], adj A =[{:(d,-b),(-c," "a):}]and |A|=ad-bc`
`A^(-1)=(adjA)/(|A|)if|A|ne 0`
So, `A^(-1)=(1)/(ad-bc)[{:(d,-b),(-c," "a):}]=(1)/(|A|)[{:(d,-b),(-c," "a):}]`
Given that `A=[{:(1+i,1+i),(-1+i,1-i):}]`
Then adj A = `[{:(1-i,-1-i),(1-i," "1+i):}]`
and, `|A|=|{:(1+i,1+i),(-1+i,1-i):}|`
`=(1-i^(2))-(i^(2)-1)=(1-i^(2)-i^(2)+1)=1+1+1+1=4`
`rArr A^(-1)=(1)/(4)[{:(1-i,-1-i),(1-i," "1+i):}]`