Evluate `int` `x^3` `dx`

Answer

A

Explanation

`because B = A -2I `
`therefore A^(-1) B = I - 2A^(-1)` ...(i)
`det [adj(I-2A^(-1)) ] = det [adj(A^(-1) B)]` [from Eq. (i)]
`= abs(adj (A^(-1)B))`
`=abs(A^(-1) B )^(2) = (abs(A^(-1)) abs(B))^(2) = (abs(B)/abs(A))^(2) ` ...(ii)
From Eq. (i), we get `B = A - 2I`
`therefore B^(3) = (A-2I)^(3) = A^(3) - 6A^(2) + 12A - 8I`
` = 5A [ because A^(3) - 6 A^(2) + 7 A - * I = 0]`
` rArr abs(B^(3))=abs(5A)`
`rArr abs(B)^(3) = 5^(3) abs(A) `
`rArr abs(B)^(3)= 5^(3) xx 8`
` rArr abs(B)^(3) = (10)^(3) `
`therefore abs(B) = 10`
From Eq. (ii), we get
`det[adj (I - 2 A ^(-1) )] = (abs(B)/abs(A))^(2) = (10/8)^(2) = 25/16`