If A and B are 2x2 matrices with det(A)=3 and det(B)=2, what is det((3A^-1)(B^T))?
1 Answer
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:
det(bb(M^(-1))) = 1/det(bb(M)) det(bb(M^(T))) = det(bb(M)) - If one row of
bb(M) is multiplied bylamda to produce a matrixbb(N) , thendet(bb(N))= lamda det(bb(M)) det(bb(M)bb(N)) = det(bb(M)) \ det(bb(N)) \ \ wherebb(M),bb(N) same dimension.
Also not that as a corollary, if we multiply a matrix by a constant,
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