Question

Given det(A) = -7 and det(B) = 8, what is the determinant of (5A^-1)*(B^T)

Answer 1

I know that the det(B^T) is the same and the det(A^-1) = 1/det(A)

so I did 5*(1/-7)*8 = 40/-7 which was not correct :c

Answer 2

B^T is fine


2 3
4 5

det = 10 - 12 = -2

inverse

5/-2 -3/-2
-4/-2 2/-2

10/4 - 12/4 = -2/4 = -1/2

the rules seem to hold

Answer 3

5k/-2 -3k/-2
-4k/-2 2k/-2

k*k = k^2 to

Answer 4

det(kA) = k^2 det(A) is what i see it at

Answer 5

but im not up to date on the rules

Answer 6

Assuming \(A\) and \(B\) are square with the same dimensions, you have
\[\det (AB)=\det A\times\det B\]
which must be the case, otherwise \(A\) wouldn't have an inverse and \(A^{-1}B^T\) wouldn't exist.