If `A=` diag (1,4,5) then `A^(-1)` is equal to

Options

A. `[(1,0,0),(0,4,0),(0,0,5)]`
B. `[(1,0,0),(0,1/4,0),(0,0,1/5)]`
C. `[(1,0,0),(0,-1/4,0),(0,0,-1/5)]`
D. None of these

Answer

B

Explanation

`A=[(1,0,0),(0,4,0),(0,0,5)]`
Now `|A|=1(20-0)=20`
and `adjA=[(20,0,0),(0,5,0),(0,0,4)]`
`:.A^(-1)=(adjA)/(|A|)=1/20[(20,0,0),(0,5,0),(0,0,4)]=[(1,0,0),(0,1/4,0),(0,0,1/5)]`
Alternate Method:
`:' "AA"^(-1)=l`
`implies[(1,0,0),(0,4,0),(0,0,5)]A^(-1)=[(1,0,0),(0,1,0),(0,0,1)]`
Applying `R_(2)to(R_(2))/4` and `R_(3)to(R_(3))/5` we get
`implies[(1,0,0),(0,1,0),(0,0,1)] A^(-1)=[(1,0,0),(0,1/4,0),(0,0,1/5)]`
`:.A^(-1)=[(1,0,0),(0,1/4,0),(0,0,1/5)]`