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Id [1//25 0x1//25]=[5 0-a5]^(-2) , then ...

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

A

`a//125`

B

`2a//125`

C

`2a//25`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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