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

1 Answer
Sep 18, 2017

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