If `A` is a non-singular matrix of order `nxxn` such that `3ABA^(-1)+A=2A^(-1)BA`, then

Options

A. `A` and `B` both are identity matrices
B. `|A+B|=0`
C. `|ABA^(-1)-A^(-1)BA|=0`
D. `A+B` is not a singular matrix

Answer

B, C

Explanation

`(b,c)` `3ABA^(-1)+A=2A^(-1)BA`
`implies3AB^(-1)+A+2A=2A^(-1)BA+2A`
`implies3A(BA^(-1)+I)=2(A^(-1)B+I)A`
`implies3A(B+IA)A^(-1)=2A^(-1)(B+AI)A`
`3A(B+A)A^(-1)=2A^(-1)(B+A)A`
Let `B+A=X`
`implies3AXA^(-1)=2A^(-1)XA`
`implies3^(n)|A||X||A^(-1)|=2^(n)|A^(-1)||X||A|`
`implies3^(n)|X|=2^(n)|X|` (as `|A|ne0`)
`implies|X|=0` or `|A+B|=0` ......`(i)`
Let `M=ABA^(-1)-A^(-1)BA`
`:.AM=A^(2)BA^(-1)-BAimpliesBA=A^(2)BA^(-1)-AM`
Now `3ABA^(-1)+A=2A^(-1)BA`
`=2A^(-1)(A^(2)BA^(-1)-AM)`
`=2ABA^(-1)-2M`
`impliesABA^(-1)+A=-2M`
`impliesA(BA^(-1)+I)=-2M`
`A(A+B)A^(-1)=-2M`
Taking determinants both sides we get
`|-2M|=|A||A+B||A^(-1)|=0`
`implies|ABA^(-1)-AB^(-1)A|=0`