We are given the following:

det(A) = -4

det(B) = 3

To solve for det(8A^-1), we first note the following:

i. The determinant of an inverse of a matrix, is just the reciprocal of the determinant of the original matrix.

ii. Scalar multiplication of a row by a constant c, also multiplies the determinant by c.

Hence, det(8A^-1) can be calculated as follows:

`det(8A^-1) = 8^4 det(A^-1) = 8^4 (1/det(A)) = 4096/-4 = -1024`

Note that we multiplied by 8^4 since A is a 4 x 4 matrix (i.e. there are 4 rows, each of which is multiplied by the scalar 8, thus, the determinant of A^-1 is multiplied to 8 for times or 8^4).

We know

`det(P^(-1))=1/det(P)`

`det(xP)=x^ndet(P)` , x is scalar and n is order of P.

Thus

`det(8A^(-1))=8^4det(A^(-1))=8^4/det(A)`

`=8^4/(det(A))`

`=8^4/(-4)=-1024`

Ans.