Question

If A and B are 2x2 matrices with det(A)=3 and det(B)=2, what is det((3A^-1)(B^T))?

Answer 1

`det((3bb(A)^-1)(bb(B)^T)) = 6`

Explanation:

We have:

`det(bb(A)) = 3`
`det(bb(B)) = 2`

We will need the following properties of determinants:

1. `det(bb(M^(-1))) = 1/det(bb(M))`
2. `det(bb(M^(T))) = det(bb(M))`
3. If one row of `bb(M)` is multiplied by `lamda` to produce a matrix `bb(N)`, then `det(bb(N))= lamda det(bb(M))`
4. `det(bb(M)bb(N)) = det(bb(M)) \ det(bb(N)) \ \` where `bb(M),bb(N)` same dimension.

Also not that as a corollary, if we multiply a matrix by a constant, `lamda`, then every element (and therefore every row) gets multiplied by that number, so that using the third property we can factor out a factor of `lamda^m` where `m` is dimension of the matrix.

So:

`det((3bb(A)^-1)(bb(B)^T)) = det(3bb(A)^-1) \ det(bb(B)^T) \ \` (property 4)
`" " = det(3bb(A)^-1) \ det(bb(B)) \ \` (property 2)
`" " = 3^2det(bb(A)^-1) \ det(bb(B)) \ \` (`A \ 2 xx2`)
`" " = 9 1/det(bb(A)) \ det(bb(B)) \ \` (property 1)
`" " = 9 xx 1/3 xx 2`
`" " = 6`