Id `[1//25 0x1//25]=[5 0-a5]^(-2)` , then the value of `x` is `a//125` b. `2a//125` c. `2a//25` d. none of these

Options

A. `a//125`
B. `2a//125`
C. `2a//25`
D. none of these

Answer

B

Explanation

Let `A=[(5,0),(-a,5)]`
`implies` adj `(A)= [(5,0),(a,5)]`
`implies A^(-1) =1/(|A|)[(5,0),(a,5)]=1/25 [(5,0),(a,5)]`
`implies A^(-2)=(A^(-1))^(2)=1/25 [(5,0),(a,5)]1/25 [(5,0),(a,5)]`
`=1/625 [(25,0),(10a, 25)]`
`=[(1/25,0),((2a)/125,1/25)]`
Now,
`[(1//25,0),(x,1//25)]=[(1/25,0),((2a)/125,1/25)]`
`implies x=2a//125`