A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`, and A( adj. A + adj. B)=B.
`AB^(-1)=`

Options

A. `B^(-1)A`
B. `AB^(-1)`
C. `A^(T)B^(-1)`
D. `B^(T)A^(-1)`

Answer

A

Explanation

`B+A=B^(2)`
`implies B^(T)B+B^(T)A=B^(T)B^(2)`
`implies I+B^(T)A=B`
`implies B^(T)A=B-I`
Now from (1) we get
`B^(2)-B=A`
`implies B(B-I)=A`
Multiplying both sides by `B^(-1)`, we get
`implies B-I=AB^(-1)`
So, from (1) and (2), we get
`AB^(-1)=B^(T)A`
`implies AB^(-1)=B^(-1)A" "( :' B B^(T)=I" then "B^(T)=B^(-1))`