Let A be a square matrix of order 3 satisfies the relation `A^(3)-6A^(2)+7A-8I=O` and `B=A-2I`. Also, det. `A=8`, then

Options

A. det. `("adj. "(I-2A^(-1))=25/16`
B. adj. `((B/2)^(-1))=B/10`
C. det. `("adj. "(I-2A^(-1)))=75/32`
D. adj. `((B/2)^(-1))=(2B)/5`

Explanation

Given that `B=A-2I` ...(i)
And `A^(3)-6A^(2)+7A-8I=O`
`:. (A-2I)^(3)=5A` ...(ii)
`:. B^(3)=5A`
`implies |B^(3)|=|5A|=5^(3)|A|=5^(3)xx8`
`implies |B|=10`
Now, from (i), we get
`:. A^(-1) B=I-2A^(-1)`
`:. |"adj. "(I-2A^(-1))|=|I-2A^(-1)|^(2)`
`=|A^(-1)B|^(2)=((|B|)/(|A|))^(2)`
`=(5/4)^(2)`
`=25/16`
adj `((B/2)^(-1))=("adj."(B/2))^(-1)=("adj."("adj." B/2))/(|"adj." B/2|)`
`=(|B/2|B/2)/(|B/2|^(2))=(B/2)/(|B/2|)=(4B)/(|B|)=(2B)/5`